Path-connectedness of Q^n

[Surlethe]

Is it even meaningful to talk about the path-connectedness of Q^n? As a subset of R^n? I was wondering about that earlier today: on the one hand, it seems that Q^n can't be path-connected because it's perforated by the reals; on the other hand, Q^n is obviously both dense in itself and in R^n, so I could "draw" a curve connecting any two points. But is it possible to construct a continuous function p(t): [0,1] -> Q^n s.t., for any two points x, y in Q^n, p(0) = x and p(1) = y? Perhaps it has to do with the fact that p does not have to be injective.

I'd appreciate any help resolving this and seeing if/why the intuitive objection that Q^n is countable while [0,1] is uncountable doesn't matter.

[Kuroneko]

The rationals ℚ are clearly disconnected, as can be easily demonstrated with A = (-inf,α)∩ℚ, B = (α,+inf)∩ℚ being open, disjoint, and covering ℚ for any irrational α. The same kind of trick can be applied to ℚ^n, with any irrational hyperplane (e.g., π_1(x) = α) partitioning ℚ^n in a similar manner. Every path-connected space is connected, so ℚ^n cannot be path-connected.

Even more strongly, for any point x in ℚ^n, [edit]we can separate x from any given y through the same hyperplane trick[/edit]. Therefore, for any x,y in ℚ^n, there is a clopen neighborhood of x that does not contain y (pick α<|x-y|)--in other words, ℚ^n is totally disconnected.

Your intuitive objection is actually very relevant, although for this particular case needlessly complex. Since ℚ^n is Hausdorff, it is path-connected iff it is arcwise-connected (an arc is a path that is also a homeomorphism). But since that would mean that there is a subset of ℚ^n homeomorphic to [0,1]; this cannot be the case because they have different cardinalities. Thus, ℚ^n is not path-connected. A related general result is the Hahn-Mazurkiewicz theorem, which states that a Hausdorff space X is a continuous image of [0,1] iff X is Peano (compact, connected, weakly locally connected, and metrizable/second-countable). I'm not going to go into the details right now simply because the two previous results are more than sufficient.

Edit: I'm not sure what kind of brain malfunction was responsible for my previous reasons for total disconnectedness, but it was incorrect. The conclusion, however, is perfectly fine.

[Winston Blake]

For once, I've learnt something from one of Kuroneko's math posts - I am now aware of the word 'clopen'.

Clopen.

CLOPEN.

CLOOOOPEN.


My math skills feel so... inadequate.

[Kuroneko]

Once again, every Hausdorff space that's path-connected is also arc-connected, which is one way to exploit cardinality differences. But earlier, you were wondering what a non-Hausdorff space might looks like. Here's an example of a path-connected space that's not arc-connected, and it make use of non-injectivity that you thought was relevant.

Take Z to be the two line segments {0,1}x[0,1] and take the quotient space X = Z/~, where ~ is the equivalence relation (a,b) ~ (c,d) iff two segments --> four-ended line segment
*-----------* {1}x[0,1]
. ^ 0'*____________*1'
. ~ | identify --> 0 * ^ *1
. v |
*-----------* {0}x[0,1] open interval (0,1)[/code]
Every open set in X that includes either 0 or 0' also includes some points in the interval, so 0 and 0' cannot be separated by open sets (same for 1 and 1'); hence X is not Hausdorff. X is path-connected, but any path from, say, 0 to 0' must "go out" on the interval at least a bit, turn around, and back-track. Therefore the path cannot be injective, and hence cannot be an arc.

[dragon]

For once, I've learnt something from one of Kuroneko's math posts - I am now aware of the word 'clopen'.

Clopen.

CLOPEN.

CLOOOOPEN.


My math skills feel so... inadequate.

Don't feel bad. I have a B.S. in math and still feel lost.

[Surlethe]

Once again, every Hausdorff space that's path-connected is also arc-connected, which is one way to exploit cardinality differences. But earlier, you were wondering what a non-Hausdorff space might looks like. Here's an example of a path-connected space that's not arc-connected, and it make use of non-injectivity that you thought was relevant.

Take Z to be the two line segments {0,1}x[0,1] and take the quotient space X = Z/~, where ~ is the equivalence relation (a,b) ~ (c,d) iff <snip>[/code]

That's pathological. It shouldn't exist.

Every open set in X that includes either 0 or 0' also includes some points in the interval, so 0 and 0' cannot be separated by open sets (same for 1 and 1'); hence X is not Hausdorff. X is path-connected, but any path from, say, 0 to 0' must "go out" on the interval at least a bit, turn around, and back-track. Therefore the path cannot be injective, and hence cannot be an arc.

Hmm. Is it valid to say that no non-Hausdorff space can be arc-connected? That doesn't seem to be true. Take the X from above, and "loop" it to make a ring with a hole in it by identifying 0 ~ 1 and 0' ~ 1'. The space is path-connected, and since you can go around the ring to get from 0 to 0', you can have an injective path connecting them. But 0 and 0' are not separable, either: they share the same neighborhood.

[Surlethe]

For once, I've learnt something from one of Kuroneko's math posts - I am now aware of the word 'clopen'.

Clopen.

CLOPEN.

CLOOOOPEN.


My math skills feel so... inadequate.

I assume it means both closed and open.

Intuitive definition of an open set: every point is an an "interior point". That is, there's always a little bit of wiggle room around a point in the set. Example: take the fractions between 0 and 1, but not including 0 and 1. If you pick any fraction between 0 and 1, you can always find another one closer to 0 or closer to 1. So if you pick 1999/2000, which is very close to (but less than) 1, you can still find another fraction -- say, 19999/20000 -- that is greater than 1999/2000 and still less than 1.

Intuitive definition of a closed set: the complement of an open set. Or, a set with an actual edge instead of fuzzily disappearing. Example: take the line of fractions and punch out (0,1) from above. The resulting set will be closed -- it will contain 0 on its "edge", and when you pick 0, you can't get any closer to 0.

So, if you followed that, a clopen set is one that is both open and closed. For example, the empty set is clopen; the surface of a sphere is clopen; and the line of real numbers is clopen.

[Kuroneko]

That's pathological. It shouldn't exist.

Despite the moral argument, four-ended line segments are simply too cute.

Hmm. Is it valid to say that no non-Hausdorff space can be arc-connected?

No. What is potentially the absolute cheapest way to construct a non-Hausdorff is to take the above line segments Z = {0,1}×[0,1] and specify the basis of the topology to be B = {{0,1}×(a,b): 0<a<b<1} (the open sets are unions of the sets in B). Now Z is just like the interval [0,1], except that that every point is "duplicated" in the sense that every open set of Z that contains the element (0,x) also contains (1,x) and vice versa. The points are set-theoretically different but not topologically distinguishable [1] from one another. It is easy to see than any distinct points of Z can be connected by an arc. Explicitly, to connect the elements (α,x) with (β,y), where α,β are either 0 or 1 and 0≤x,y,≤1), we need only to specify the function f:[0,1]→Z by f(0) = (α,x), f(w) = (0,x+w(y-x)) on 0<w<1, f(1) = (β,y).

Perhaps a bit more intuitively that the above, all we're doing is taking the interval [0,1], and for each x in [0,1], inventing a completely new element x', and sticking x' into every open in which x appears.

[1] Thus, this space is actually not even T0 (Kolmogorov), much less T1 (Frechet) or T2 (Hausdorff). For any points x,y in X, X is T0 iff there exists an open set containing x but not y or vice versa (topological distinguishability); it is T1 is there is a pair of open sets, one containing x but not y and the other y but not x; it is T2 iff there is a disjoint such pair. It fairly easy to construct a space that's T0 but not T2 and arc-connected using a similar method to the above. Getting a space that's T1 but not T2 and arc-connected may be a bit trickier.

That doesn't seem to be true. Take the X from above, and "loop" it to make a ring with a hole in it by identifying 0 ~ 1 and 0' ~ 1'. The space is path-connected, and since you can go around the ring to get from 0 to 0', you can have an injective path connecting them.

It's not really a hole--there is extra point rather than a lack of it. But there's something a bit fishy about this particular example. An arc is a path that is also homeomorphism to its range (i.e., a path with a continuous inverse function). In this case, the range of the arc is the entire space, so we must have this kind of 'extra-point circle' homeomorphic to [0,1]. This means that the respective spaces have the same topological properties, and since [0,1] is Hausdorff, so must this pseudo-circle.

Let's see what went wrong. First, is the ring really non-Hausdorff? Well, starting back from the two line segments, we have the following examples of open sets: A = {0}×[0,ε), B = {1}×(1-ε,1]--some initial open part of the bottom segment and some final open part of the upper one. When the intermediate points were identified, we not have A' = [0,ε) and B' (1-ε,1'], where A' does not include 0' and B' does not include 1. Now, with your final identification 0~1 and 0'∼1', we still have the same two open sets A' and B', except that now 1' is the same as 0'.

Therefore, after your identification, 0 and 0' can actually be separated by disjoint open sets, and your ring is actually Hausdorff. It's somewhat counter-intuitive because in some sense the "opposite" endpoints were identified, but the identification still leaves two endpoints, so it's not completely surprising that what we get now is just a regular two-ended line segment.

Intuitive definition of an open set: every point is an an "interior point". That is, there's always a little bit of wiggle room around a point in the set. Example: take the fractions between 0 and 1, but not including 0 and 1. If you pick any fraction between 0 and 1, you can always find another one closer to 0 or closer to 1.

That's true, but the importance of being open is not so much the fact that you can find such elements (which is optional), but that for every one you do find, it will be in the open set. In other words, is G is an open set, then for any x in G, anything within some small distance ε>0 is also in G (for some sufficiently small ε). [This definition is only valid in metric spaces, but since both R^n and Q^n are metric spaces, it is quite applicable.]

In the reals, open sets are just unions of open intervals, and a set G is open iff around each element of G, there is an open interval contained entirely in G.

Intuitive definition of a closed set: the complement of an open set. Or, a set with an actual edge instead of fuzzily disappearing. Example: take the line of fractions and punch out (0,1) from above. The resulting set will be closed -- it will contain 0 on its "edge", and when you pick 0, you can't get any closer to 0.

That's a good way to think about closed sets. Put another way, a set F is closed iff for every convergent sequence composed of elements of F, the limit of that sequence is also in F. So, for example, in the reals, [0,1) is not closed because one the sequence 1-1/2^n has a limit of 1, but 1 is not in that set. On the other hand, [0,1] is closed (and is in fact the complement of the open set (-inf,0)U(1,+inf)).

In the reals, closed sets are intersections of closed intervals, and a set F is closed iff around each element not in F, there is an open interval completely disjoint with F.

Speaking of closed sets, the following definitions of the Hausdorff (T2) property are equivalent:
1. For any distinct x,y, there are disjoint open sets U,V with x in U, y in V.
2. For any x, the intersection of all closed sets that contain x is precisely {x}. (In the above example, the intersection of all closed sets that contain 0 is actually {0,0'}, etc.)

So, if you followed that, a clopen set is one that is both open and closed. For example, the empty set is clopen; the surface of a sphere is clopen; and the line of real numbers is clopen.

In itself, yes. The surface of a sphere is not clopen in R^3. There is actually a good alternative definition of connectedness:
A space X is connected iff the only clopen sets in X are the empty set and X itself.
It's by virtue of the fact that we can find nontrivial clopen sets in the rationals (such (-inf,α)∩Q for any irrational α: it is open because it is an open interval, and its complement [α,+inf)∩Q is the same as (α,+inf)∩Q, also an open interval because α is irrational) that the rationals are disconnected. Being disconnected, they cannot be path-connected.


Quote of the Week: "The complement is the intersection of the complement with the entire space."
(Definition of complement; said on the first day of a real analysis class while reviewing set theory.)

[Ariphaos]

... Every open set in X that includes either 0 or 0' also includes some points in the interval, so 0 and 0' cannot be separated by open sets (same for 1 and 1'); ...

...what the hell!?

My vote is with Surlethe >_>

[Jaepheth]

Real Analysis isn't my strong point... but wouldn't Q^n not be path connected since the elements are countable (i.e. Lebesgue measure of 0)?

[Surlethe]

Real Analysis isn't my strong point... but wouldn't Q^n not be path connected since the elements are countable (i.e. Lebesgue measure of 0)?

I'm not terribly familiar with real analysis. Why would having a Lebesgue measure of 0 imply no path-connectedness?

[Glass Pearl Player]

Real Analysis isn't my strong point... but wouldn't Q^n not be path connected since the elements are countable (i.e. Lebesgue measure of 0)?

I think you want to use the 1-dimensional Haussdorf measure H^1. The H^1- measure of a path in Q^n - if such a thing would exist - should be at least the distance of its endpoints, the H^1-measure of Q^n is zero. But a subset cannot exceed the measure of its superset, so we have a contradiction.

The essence of the argument is thus: Q^n has Haussdorf dimension 0, while a path has Haussdorf dimension 1, and thus can't fit into Q^n.

[Surlethe]

That's pathological. It shouldn't exist.

Despite the moral argument, four-ended line segments are simply too cute.

Cute though they may be, where do you draw the line? Five-ended line segments? Six-ended? n-ended? Aleph_1-ended?!

Hmm. Is it valid to say that no non-Hausdorff space can be arc-connected?

No. What is potentially the absolute cheapest way to construct a non-Hausdorff is to take the above line segments Z = {0,1}×[0,1] and specify the basis of the topology to be B = {{0,1}×(a,b): 0<a<b<1} (the open sets are unions of the sets in B).

Can the arc-connectedness vary based on the given topology?

Now Z is just like the interval [0,1], except that that every point is "duplicated" in the sense that every open set of Z that contains the element (0,x) also contains (1,x) and vice versa. The points are set-theoretically different but not topologically distinguishable [1] from one another. It is easy to see than any distinct points of Z can be connected by an arc. Explicitly, to connect the elements (α,x) with (β,y), where α,β are either 0 or 1 and 0≤x,y,≤1), we need only to specify the function f:[0,1]→Z by f(0) = (α,x), f(w) = (0,x+w(y-x)) on 0<w<1, f(1) = (β,y).

Perhaps a bit more intuitively that the above, all we're doing is taking the interval [0,1], and for each x in [0,1], inventing a completely new element x', and sticking x' into every open in which x appears.

This all makes sense. Since (0,x) and (1,x), for 0≤x≤1, are topologically indistinguishable, f can jump between them at will and maintain its continuity. Cute.

[1] Thus, this space is actually not even T0 (Kolmogorov), much less T1 (Frechet) or T2 (Hausdorff). For any points x,y in X, X is T0 iff there exists an open set containing x but not y or vice versa (topological distinguishability); it is T1 is there is a pair of open sets, one containing x but not y and the other y but not x; it is T2 iff there is a disjoint such pair. It fairly easy to construct a space that's T0 but not T2 and arc-connected using a similar method to the above. Getting a space that's T1 but not T2 and arc-connected may be a bit trickier.

Hmm. Let T be such a space. Requirements:

1. for all x,y∈T, there exist X∋x and Y∋y s.t. Y∌x and X∌y;
2. For all X and Y as above, X∩Y≠∅;
3. There exists homeomorphic f:[0,1]→T s.t. f(0)=x and f(1)=y ∀x,y∈T.

Perhaps take a simple ring and choose the topology so that it satisfies 1 and 2? I'm at a loss how to do it.

That doesn't seem to be true. Take the X from above, and "loop" it to make a ring with a hole in it by identifying 0 ~ 1 and 0' ~ 1'. The space is path-connected, and since you can go around the ring to get from 0 to 0', you can have an injective path connecting them.

It's not really a hole--there is extra point rather than a lack of it. But there's something a bit fishy about this particular example. An arc is a path that is also homeomorphism to its range (i.e., a path with a continuous inverse function). In this case, the range of the arc is the entire space, so we must have this kind of 'extra-point circle' homeomorphic to [0,1]. This means that the respective spaces have the same topological properties, and since [0,1] is Hausdorff, so must this pseudo-circle.

Let's see what went wrong. First, is the ring really non-Hausdorff? Well, starting back from the two line segments, we have the following examples of open sets: A = {0}×[0,ε), B = {1}×(1-ε,1]--some initial open part of the bottom segment and some final open part of the upper one. When the intermediate points were identified, we not have A' = [0,ε) and B' (1-ε,1'], where A' does not include 0' and B' does not include 1. Now, with your final identification 0~1 and 0'∼1', we still have the same two open sets A' and B', except that now 1' is the same as 0'.

After the identification, A' and B' are no longer open sets, are they? That is, no neighborhood around 0 is contained in A' and no neighborhood around 0' is contained in B'.

Therefore, after your identification, 0 and 0' can actually be separated by disjoint open sets, and your ring is actually Hausdorff. It's somewhat counter-intuitive because in some sense the "opposite" endpoints were identified, but the identification still leaves two endpoints, so it's not completely surprising that what we get now is just a regular two-ended line segment.

I was thinking that it would look less like an identification with two endpoints and more "like" this.

Intuitive definition of a closed set: the complement of an open set. Or, a set with an actual edge instead of fuzzily disappearing. Example: take the line of fractions and punch out (0,1) from above. The resulting set will be closed -- it will contain 0 on its "edge", and when you pick 0, you can't get any closer to 0.

That's a good way to think about closed sets. Put another way, a set F is closed iff for every convergent sequence composed of elements of F, the limit of that sequence is also in F.

Ahh, every convergent sequence. Important detail, that. I was about to ask whether [0,inf) was closed.

So, for example, in the reals, [0,1) is not closed because one the sequence 1-1/2^n has a limit of 1, but 1 is not in that set. On the other hand, [0,1] is closed (and is in fact the complement of the open set (-inf,0)U(1,+inf)).

In the reals, closed sets are intersections of closed intervals, and a set F is closed iff around each element not in F, there is an open interval completely disjoint with F.

Finite intersections, right?

Speaking of closed sets, the following definitions of the Hausdorff (T2) property are equivalent:
1. For any distinct x,y, there are disjoint open sets U,V with x in U, y in V.
2. For any x, the intersection of all closed sets that contain x is precisely {x}. (In the above example, the intersection of all closed sets that contain 0 is actually {0,0'}, etc.)

As an exercise, proof. Let T be the given (non-trivial) space.

(→) Suppose 1. Then, for any closed X_αy containing x, with X_αy≠T, X_y⊂(T/V_α), with y∈V for some y and open V_α. In fact, this will hold for every y≠x. So consider A=∪_{y≠x}(∪_{α}X_αy). Suppose now for contradiction that A={x}∪K (a disjoint union), with K nonempty (A is trivially nonempty since x is by construction an element of each X_y). Then K will include as an element some y'. But this cannot be: this would imply that every X_αy contained y', including X_αy', which cannot. Contradiction.

(←) Suppose 2. Suppose further that y≠x are both elements of T. The union of every open set that does not contain x is T/{x}; likewise for y. Since y∈T/{x} and x∈T/{y}, there must exist open sets U,V s.t. x∈U∌y and y∈V∌x. (This shows T is Frechet - T1). Now suppose for contradiction there are no disjoint U,V that satisfy this condition. ... I'm having a bit of trouble finishing this proof off; it seems it could possibly be the case that for every such U,V, their intersection is a different element of T, thus rendering a proof down to contradiction much harder to see; and I'm at a loss how to actually construct disjoint U and V without assuming nice things like a sense of distance.

So, if you followed that, a clopen set is one that is both open and closed. For example, the empty set is clopen; the surface of a sphere is clopen; and the line of real numbers is clopen.

In itself, yes. The surface of a sphere is not clopen in R^3. There is actually a good alternative definition of connectedness:
A space X is connected iff the only clopen sets in X are the empty set and X itself.

I think the proof of this is beyond me at the moment, but that makes some sense.

Quote of the Week: "The complement is the intersection of the complement with the entire space."
(Definition of complement; said on the first day of a real analysis class while reviewing set theory.)

[Kuroneko]

Real Analysis isn't my strong point... but wouldn't Q^n not be path connected since the elements are countable (i.e. Lebesgue measure of 0)?

As mentioned before, countability is relevant because in Hausdorff spaces, path-connected is equivalent to arc-connected, and we cannot have arcs in countable spaces. But Lebesgue measure 0 is relevant only for n = 1.

Glass Pearl Player's suggestion is very nice.

I'm not terribly familiar with real analysis. Why would having a Lebesgue measure of 0 imply no path-connectedness?

It doesn't. For n>1, any hyperplane of R^n has Lebesgue measure zero and is clearly path-connected. However, for n=1, any null set will contain no intervals and thus cannot be connected.

---

Cute though they may be, where do you draw the line? Five-ended line segments? Six-ended? n-ended? Aleph_1-ended?!

Any lines you draw can be made to have split ends. Just take ω_2×[0,1] any apply the same equivalence relation for Aleph_2-ended, for example.

Can the arc-connectedness vary based on the given topology?

Absolutely. The topology defines the open sets, so of course it matters quite a bit. If Q^n was given the discrete topology {φ,Q^n}, it would obviously be path-connected: every function to a space with a discrete topology is continuous. However, in this case it would not be Hausdorff. Arc-connectedness has a similar dependence on topology, as does every other kind of connectedness.

All of the above discussion implicitly assumed that the topology is the one inherited from Q^n being a subspace of R^n, or equivalently in the standard manner of a Euclidean metric.

Hmm. Let T be such a space. Requirements:
for all x,y∈T, there exist X∋x and Y∋y s.t. Y∌x and X∌y;
For all X and Y as above, X∩Y≠∅;
There exists homeomorphic f:[0,1]→T s.t. f(0)=x and f(1)=y ∀x,y∈T.

Close, but not quite. The first one's fine, but...
(1') [T1] for all x,y∈T, there exist neighborhoods X,Y of x,y (resp.) s.t. y is not in X and x is not in Y. If we let N_x denote an arbitrary neighborhood [1] of x, then (∀x,y∈T)(∃N_x,N_y)(x∉N_y∧y∉N_x).
(2') [not T2] for some x,y∈T, no pair of neighborhoods N_x,N_y is disjoint: (∃x,y)(∀N_x,N_y)(N_x∩N_y≠∅).
(3') for every x,y∈T, there exists f:[0,1]→T s.t. f(0)=x, f(1)=1), and f is a homeomorphism between [0,1] and f([0,1]). Or equivalently,
(3') for every x,y∈T, there exists a continuous open [2] injection f:[0,1]→T s.t. f(0)=x, f(1)=y.


[1] There are two different definitions of "neighborhood", but the difference doesn't matter at all here. Either (1) a neighborhood is an open set that contains x or (2) a set that contains another open set that in turn contains x. The latter definition is somewhat more general and useful in other contexts, but here we can just assume that all neighborhoods are themselves open. It changes absolutely nothing here.
[2] An open function maps open sets to open sets, i.e., for every open G, f(G) is open. A continuous function is one for which the inverse f' (which is not necessarily a function) maps open sets to open sets.

Perhaps take a simple ring and choose the topology so that it satisfies 1 and 2? I'm at a loss how to do it.

I'm not certain, but correcting an mistake I'll get into later, the minimal requirement for T1 is that every singleton is closed, which gives the cofinite topology (open sets are complements of finite ones). Perhaps having a Cartesian product of a well-behaved space with one that has a cofinite topology would work, e.g., X = R²×[0,1], but with [0,1] having said cofinite topology. I haven't checked it, though.

After the identification, A' and B' are no longer open sets, are they? That is, no neighborhood around 0 is contained in A' and no neighborhood around 0' is contained in B'.

Aa. Yes, you're right. This means that your ring is not arc-connected. It seems visual intuition failed us both. The space is not Hausdorff, and I apologize for my mistake.

To constract with the standard line segment [0,1], if you identify 0~1 here, then we have [0,1) under a different topology. In the new topology, the open set [0,ε) = {0}∪(0,ε) corresponds to {0,1}∪(0,ε) in the old one, which is not open. However, [0,ε)∪(1-δ,1) corresponds to {0,1}∪(0,ε)∪(1-δ,1) = [0,ε)∪(1-δ,1], which is open in [0,1]. All of this is just a sanity test of the quotient topology: if we identify 0 and 1 on [0,1], the space should "wrap around" into a circle. And that's exactly what happens: we do not see open sets in the form [0,ε) anymore after the identification, but we do see open sets with elements "to both sides" of 0 (which we can call a north pole if we wish).

Now, in the four-ended line segment, your identification is 0~1, 0'~1' as you had. The openness of A' = {0}∪(0,ε) is equivalent to the openness of {0,1}∪(0,ε) in the original four-ended line segment, and since it is a quotient space of two line segments {0,1}×[0,1], it in turn is equivalent to the openness of {(0,0),(0,1)}.∪.{0,1}×(0,ε). This is not open.

A continuous function f is one whose inverse f' (which is not necessarily a function itself) is open, i.e., maps open sets to open sets: for all open G, f'(G) is open. Since a homeomorphism is a continuous function with a continuous inverse function, we can characterize an arc as an open injective path. The two 'endpoints' of your ring-like space X (which we can call 0 and 0'), do actually have a path between them; in fact there are two interesting ones:
f:[0,1]→X by p(0) = 0, p(w) = w [0<w<1], p(1) = 0'
g:[0,1]→X by p(0) = 0, p(w) = 0 [0<w<1], p(1) = 0'
Yes, the latter is actually continuous, because for any open set in X, g'(X) is empty, or [0,1), or [0,1], all of which are open in [0,1]. There are plenty of others that go up to some point and loop back, of course, but being non-injective, they obviously can't be arcs (neither can g).

Now, f is a path, but is it an open path? Since [0,ε) is open in [0,1], f({0}∪(0,ε)) = {0}∪(0,ε) should be open in X. But as shown above, it is not. Hence f is not open and thus cannot be an arc.

Ahh, every convergent sequence. Important detail, that. I was about to ask whether [0,inf) was closed.

It actually is. And it makes sense, too, being the complement of the open interval (-inf,0).

In the reals, closed sets are intersections of closed intervals, and a set F is closed iff around each element not in F, there is an open interval completely disjoint with F

Finite intersections, right?

No, arbitrary. Arbitrary unions of open sets are open, so by de Morgan's law, arbitrary intersections of closed sets are closed. They're dual to one another in this manner. On the other hand, finite intersections of open sets have to be open, and so only finite unions of closed sets are guaranteed to be closed.

As an exercise, proof. Let T be the given (non-trivial) space.

Ack. It should have been T1. For T2, take intersections of the closures of neighborhoods.

(→) Suppose 1. Then, for any closed X_αy containing x, with X_αy≠T, X_y⊂(T/V_α), with y∈V for some y and open V_α. In fact, this will hold for every y≠x. So consider A=∪_{y≠x}(∪_{α}X_αy).

Er, I assume the union is a typo, and that you mean A is the intersection X_αy ranging over all closed sets containing x but not y. (As something worth emphasizing, of course A is itself closed, being an intersection of closed sets.)

Suppose now for contradiction that A={x}∪K (a disjoint union), with K nonempty (A is trivially nonempty since x is by construction an element of each X_y). Then K will include as an element some y'. But this cannot be: this would imply that every X_αy contained y', including X_αy', which cannot. Contradiction.

Excellent. Notice that this direction works for T1 just as well as T2 spaces. All we really needed is to get at least one open V_α around y not containing x, for each fixed y in T.

(←) Suppose 2. Suppose further that y≠x are both elements of T. The union of every open set that does not contain x is T/{x}; likewise for y. Since y∈T/{x} and x∈T/{y}, there must exist open sets U,V s.t. x∈U∌y and y∈V∌x. (This shows T is Frechet - T1).

Right. Or more directly, since {x} is closed, T\{x} is open, and similarly for T\{y}. (As a notational aside, the forward slash usually denotes a quotient space for sets; set difference is either minus or backslash.)

I'm having a bit of trouble finishing this proof off ...

That part's completely my fault. For T2, the intersection should range not over all closed sets but just ones that are the closures of open sets containing x. (Interestingly, T1 is also equivalent to the intersection all open G_x∋x being precisely {x}.)

A space X is connected iff the only clopen sets in X are the empty set and X itself.

I think the proof of this is beyond me at the moment, but that makes some sense.

Oh, it most certainly is not. One doesn't need any topology at all or any properties of open sets; just set theory and two definitions:
(Def) A set is closed iff its complement is open; a set is clopen iff it is both open and closed.
(Def) X is connected iff there do not exist nonempty open disjoint A,B with A∪B = X.

[Surlethe]

I unfortunately haven't the time tonight to respond to the entirety of the latest reply, but I do have this interesting and relevent observation regarding an arc-connected T1 space that is not Hausdorff, thanks to my topology professor.

Aha, I see. I did not read your e-mail carefully enough. Too much multitasking, I suppose.

So, you want your space to be ARC-connected and not only PATH-connected. That is possible as well:

EXAMPLE

Let X be the real line, but with fewer open sets: from the collection of all subsets of the real line which are usually open, keep only those which contain some ray (-infinity,a) to the left AND some ray (b,+infinity) to the right; also keep the empty set.

1. Verify that this indeed defines a topology on X.

2. You will quickly see that X is T1 but not T2.

3. To see that X is arc-connected, observe that for every closed subinterval [c,d] the subspace topology as a subset of X agrees with the subspace topology as a subset of the usual real line.

If I understand correctly the topology you have defined,

1. The collection T = {U in X: (-inf,a) and (b,inf) are in U for some real a,b} is a topology on the real line X.
   • Empty set and X are trivially in T, since X is a "usually" open interval (-inf, inf) and thus contains the prescribed rays, while the empty set is in the topology by fiat .
   • Say K and L are (non-empty) elements of T. Then K U L will also be in T: the requisite rays will absorb each other, to put it inaccurately. More to the point, K will contain (-inf,a) and (b,inf), while L will contain (-inf,c) and (d,inf). K U L will therefore contain (-inf,max[a,c]) and (min[b,d],inf), and is an element of the topology. The case of any unions follows immediately by the closure of the real line, I believe. -- i.e., max[(a,b)] = b exists. The case where K and L are either or both empty is trivial.
   • Say M and N are (non-empty) elements of T. As above, M and N will contain (-inf,a) and (b,inf), and (-inf,c) and (d,inf) respectively. M intersect N will contain (-inf,min[a,c]) and (max[b,d],inf), which is again in T (although it is non-empty; this will be important). The case of arbitrary intersections follows similarly by induction. The case where M and L are either or both empty is trivial.
2. X is T1 but not T2. Use the following definition of "neighborhood": a set N is a neighborhood of x in X iff it is an open set containing x.
   • X is T1. Say x and y are in X, and WLOG say x>y; then choose as a neighborhood of x the set (-inf,y-e])U(x-e,inf), and as a neighborhood of y, choose (-inf,y+e)U(x,inf), where e<x-y. Clearly, the neighborhood of x does not contain y, and vice-versa; X is thus T1.
   • X is not T2. Given x and y, no two neighborhoods can be disjoint; they are connected at infinity. More accurately, from above, the intersection of any two non-empty open sets is again non-empty; since neighborhoods must necessarily contain some elements, they are non-empty and so X is not Hausdorff.
3. Consider [a,b] in X. The subspace topology T on [a,b] is then {Y intersect [a,b]:Y contains (-inf,c) and (d,inf) for real c,d} -- that is, {(c,d):a≤c,d≤b}U{[a,d):a<d≤b}U{(c,b]:a≤c<b}U[c,d]Uempty set. This, of course, agrees with the subspace topology on [0,1] in the usual topology on the reals (see p. 90 in the textbook) by replacing a with 0 and b with 1. So construct a function f:[0,1]->[a,b] by f(t) = (b-a)t+a. Open sets are mapped to open sets; f is continuous. Moreover, f is linear and therefore has a continuous inverse; it is a homeomorphism. (Alternately, construct g:[a,b]->[a,b] by the identity map for an even more trivial homeomorphism). I am not sure why it is important to observe that the subspace topology on [a,b] in X agrees exactly with that of [a,b] in the usual topology on the reals.

This is cute. I hadn't thought to link the neighborhoods at infinity ("far away") instead of somewhere "between" x and y.

[Kuroneko]

Oh, I had almost forgotten about this. If you want another example, my previous hunch also works and is fairly intuitive. Let E be the reals with the Euclidean topology (basis [1]: ℬ_E = {(a,b): a,b in R}) and F be the reals with cofinite topology (basis: ℬ_F = finite intersection of {(-∞,x)U(x,+∞): x in R}). Then clearly F is T1 but not T2 [2], so X = E×E×F with the product topology [3] is T1 (actually, the product of T1 spaces is always T1).

It is easy to see that X is not T2. For example, in trying to separate the points x = (0,0,x') and y = (0,0,y'), for example, every neighborhood of x is going to contain a subset in the form N_x = (-ε,ε)×(-δ,δ)×U and every neighborhood of y will contain an N_y = (-ε,ε)×(-δ,δ)×V, where U,V are open neighborhoods of 0,1 (resp.) in F, for some sufficiently small ε,δ>0. But then if N_x, N_y are disjoint, it must be the case that U,V are disjoint neighborhoods of x',y' in F. This is impossible because F is not T2.

However, X is also arc-connected: in fact, any path for which any particular value of the third coordinate (z) occurs at most once will also be an arc. I think it makes in the end this example is very intuitive: we make a non-T2 space by crossing a standard plane with F, and since we have two dimensions to play in, we can easily afford to make paths which do not have take any value of the third coordinate more than once, thus hiding the non-T2 nature of F. That's why E²×F is needed rather than E×F in this example--to make these sorts of "detours" trivial between any two points with different z-coordinates. For points with the same z-coordinates, making an arc is even easierl [4].

Your professor's example is very pretty. The topology is such that a set is open iff it is an open neighborhood of infinity in the one-point compactification of the reals (think 'Riemann circle' with the top point being infinity), but of course with infinity being removed. I wish I'd thought of that.


[1] A set is open iff it is an arbitrary union of members of the basis.
[2] It is in fact the minimal topology on the reals which is still T1, in the sense that every open set in F is also open in every T1 topology on the reals.
[3] Product topology basis: {G_E×G'_E×G_F: G_E,G'_E open in E, G_F open in F}. In other words, a set G is open in X iff it is an arbitrary union of products of two sets open in E and an open set in F.
[4] If the two points (x,y,z), (x',y',z) have the same z-coordinate, then we have an arc f':[0,1]→E² between (x,y) and (x',y') and just define a new arc f:[0,1]→X by f(w) = f'(w)×{z}.
Edit: intersection condition

[Kuroneko]

3. ... So construct a function f:[0,1]->[a,b] by f(t) = (b-a)t+a. Open sets are mapped to open sets; f is continuous.

Open sets being mapped to open sets means that the inverse is continuous (and a continuous function if it is a function in the first place).

Moreover, f is linear and therefore has a continuous inverse; ...

It has a continuous inverse because f is open (maps open sets to open sets). But you haven't shown that f itself is continuous, although that's easy to do.

I am not sure why it is important to observe that the subspace topology on [a,b] in X agrees exactly with that of [a,b] in the usual topology on the reals.

Because if you know that, then f and its inverse being a continuous is trivial, from the fact they are both linear functions (in the geometric sense). Otherwise, you should compute the inverse f'(s) = (s-a)/(b-a) and explicitly verify that f' is open (which would mean that f is continuous).

I say this because your phrasing seems to imply that any real function in the form f(x) = mx+b is automatically bicontinuous. If the topology is allowed to vary, it is not the case. Whether or not it is continuous depends on the topology; in the standard Euclidean topology, of course the implication is correct. That's why noting that the subspace topology is identical to the Euclidean topology is important, since it allows you to use the results about continuity that you learned in basic calculus.

[Surlethe]

3. ... So construct a function f:[0,1]->[a,b] by f(t) = (b-a)t+a. Open sets are mapped to open sets; f is continuous.

Open sets being mapped to open sets means that the inverse is continuous (and a continuous function if it is a function in the first place).

Ah right, duh. I should have seen that.

Moreover, f is linear and therefore has a continuous inverse; ...

It has a continuous inverse because f is open (maps open sets to open sets). But you haven't shown that f itself is continuous, although that's easy to do.

Ahh, I see. The fact that the subspace topologies agrees exactly means that neighborhoods are mapped to by neighborhoods, and so f is continuous. The fact it's an open map means (as per above) that it will have a continuous inverse, and so is a homeomorphism.

I am not sure why it is important to observe that the subspace topology on [a,b] in X agrees exactly with that of [a,b] in the usual topology on the reals.

Because if you know that, then f and its inverse being a continuous is trivial, from the fact they are both linear functions (in the geometric sense).

It seems that for any subset Y of X that is not in every neighborhood of infinity, Y has a subspace topology from (X,T) that exactly agrees with the usual. Is it correct to say that X is exactly the same as the reals under the ordinary topology when infinity is not involved?

In any case, as an (easy) exercise, say the subspace topologies of [a,b] under the ordinary topology and under T are the same. Take the identity map between the two, and of course the preimage of an open set will be open.

Otherwise, you should compute the inverse f'(s) = (s-a)/(b-a) and explicitly verify that f' is open (which would mean that f is continuous).

That seems rather tedious.

I say this because your phrasing seems to imply that any real function in the form f(x) = mx+b is automatically bicontinuous.

Is that like being bicurious?

If the topology is allowed to vary, it is not the case. Whether or not it is continuous depends on the topology; in the standard Euclidean topology, of course the implication is correct. That's why noting that the subspace topology is identical to the Euclidean topology is important, since it allows you to use the results about continuity that you learned in basic calculus.

Ah, that makes sense. If, say, X were the reals with the trivial topology, then there's no point in talking about a continuous function at all.

[Kuroneko]

It seems that for any subset Y of X that is not in every neighborhood of infinity, Y has a subspace topology from (X,T) that exactly agrees with the usual. Is it correct to say that X is exactly the same as the reals under the ordinary topology when infinity is not involved?

Er... that depends on what you mean. Interpreted literally ("in" = "subset of"), there is no subset Y of X that is not contained in every neighborhood of infinity, so the statement is vacuously true. If by "not in every neighborhood", however, you mean "disjoint with some neighborhood", then that would be correct. Equivalently, the subspace topology for any bounded subset Y of X is exactly the Euclidean topology [1].

Is that like being bicurious?

Well, it means not only being open to both ways, but also being invertible--and that would be quite something to see!

Ah, that makes sense. If, say, X were the reals with the trivial topology, then there's no point in talking about a continuous function at all.

Right.


[1] Technically, we shouldn't be talking about boundedness in X because X is not a metric space, but in this case the meaning is obvious.