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As the title said, I want to show that an uncountable finite complement space is path-connected. I found that this question is answered in here. However, I'm having trouble understanding its proof. I'll use the same notation conventions defined there.

Let $T=(S,\tau)$ be a finite complement topology on a uncountable set $S$. We want to prove that $T$ is path-connected. Let $a,b\in S$ such that $a\ne b$.

My questions are as follows:

  1. Why $a$ and $b$ are contained in a subset $X\subseteq S$ whose cardinality is the same as that of $[0..1]$?.
  2. Why a bijection $f:[0,1] \to S$ such that $f(0)=a$ and $f(1)=b$ is continuous?

Any help would be much appreciated.

user
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2 Answers2

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Assuming the continuum hypothesis, the cardinality of R is the least of uncountable sets which implies that every uncountable set has cardinality larger than or equal to the cardinality of R, thus it has a subset that has the same cardinality as R. The second question, let U be open in X (not S), then U has a finite complement, the pre-image of a finite set by a bijection f is also finite, and every finite set is closed in R, now $f^{-1}(U) = [0,1] - f^{-1}(U)$, thus $f^{-1}(U)$ is open [0,1], thus f is continuous.

Burrrrb
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The result at that link is assuming the Continuum Hypothesis. Which seems a little silly; the real result is that any finite-complement space of cardinality at least $c$ is arc-connected, CH only comes in showing that uncountable implies cardinality at least $c$.

You should also note that an arc is an injective path.

Proposition If $X$ is a finite-complement space and $f:[0,1]\to X$ is injective then $X$ is continuous.

Proof: If $S\subset X$ is open then $[0,1]\setminus f^{-1}(S)$ is finite, hence $f^{-1}(S)$ is open. QED.

Cor If $X$ is a finite-complement space of cardinality at least $c$ then $X$ is arc connected.

Proof: Saying $X$ has cardinality at least $c$ says precisely that there is an injective map $f:[0,1]\to X$. So there is an arc from $f(0)$ to $f(1)$. But any bijection from $X$ to itself is a homeomorphism; hence there is an arc joining any two points.

David C. Ullrich
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  • Can you explain why "saying $X$ has cardinality at least $c$ says precisely that there is an injective map $f:[0,1]\to X$"? – user Jul 04 '16 at 16:11
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    @user This is a basic set theory thing; in fact, given that $[0,1]$ has cardinality $c$, saying there is such an injection is actually the definition of "$X$ has cardinality at least $c$". Probably there's a lot you need to learn about this sort of thing; you might start at https://en.wikipedia.org/wiki/Cardinal_number – David C. Ullrich Jul 04 '16 at 16:23
  • I see, for instance, because of http://math.stackexchange.com/questions/660997/proving-that-cardinality-of-the-reals-cardinality-of-0-1. My next question is why "any bijection from $X$ to itself is a homeomorphism"? Did you use the fact that $X$ is compact here? – user Jul 04 '16 at 16:39
  • @user No, that's because $X$ has the finite-complement topology. Think about it, using the definition of "continuous"... – David C. Ullrich Jul 04 '16 at 16:50
  • @user .$c$, the cardinal of the reals, is also the cardinal of $[0,1]$....The def'n of "The cardinal of $ X $ is greater than or equal to the cardinal of $ Y $ " is "There exists an injection $f:Y\to X. $" – DanielWainfleet Jul 04 '16 at 16:53
  • I noticed that if X is finite or countable and has at least $2$ points then $X$ is not path-connected . What can we say if we assume $X$ is uncountable and $ |X|<c $ ? – DanielWainfleet Jul 04 '16 at 16:57
  • Let $g$ be any bijection from $X$ to itself. For every $U \subseteq X$ is open then $g(U^c)$ and $g^{-1}(U^c)$ are finite because $U^c$ is finite. Since $g(U)^c=g(U^c)$ and $g^{-1}(U^c)=g^{-1}(U)^c$, $g(U)^c$ and $g^{-1}(U)^c$ all are finite i.e. $g(U), g^{-1}(U)$ are open. It follows that, $g$ is a homeomorphism. Does this look good? – user Jul 04 '16 at 17:04
  • @user254665 great – David C. Ullrich Jul 04 '16 at 17:09
  • @user You're way ahead of me on this one. Ok, $f:[0,1]\to X$ is continuous if and only if the inverse image of every singleton is closed, so $X$ is path-connected if and only if $[0,1]$ is the union of at least two but no more than $|X|$ disjoint closed sets. I don't even see why $[0,1]$ is not the union of countably many disjoint closed sets. I imagine that's obvious? – David C. Ullrich Jul 04 '16 at 17:20
  • @user . Re your most recent comment on a bijection on $X , $ it's good, although you should say " for every open $ U\subset X $ with $ U\ne \emptyset $." But is it related to the Q at hand? – DanielWainfleet Jul 04 '16 at 17:23
  • @user254665 If you look at the proof I gave it's relevant to that: Started with a continuous injective $f:[0,1]\to X$, deduced that there was an arc from $f(0)$ to $f(1)$, and then got arc-connectdeness from that homogeneity – David C. Ullrich Jul 04 '16 at 17:30
  • @user254665 So anyway, why is $[0,1]$ not the union of countably many disjoint closed sets? Go ahead and make me feel stupid, I can take it... – David C. Ullrich Jul 04 '16 at 17:36
  • @user254665 Never mind, I found it. Haven't had a chance to look at it. If I were wondering about your question about $X$ uncountable of cardinality less than $c$ I'd start there, looking to see whether the "uncountable" that's proved might actually be $c$. http://math.stackexchange.com/questions/6314/is-0-1-a-countable-disjoint-union-of-closed-sets – David C. Ullrich Jul 04 '16 at 18:02
  • @DavidC.Ullrich Yes, you're right. I should have better said "for every open $U\subseteq X$ with $U\ne \emptyset$." For $U=\emptyset$ we have $g(\emptyset)=g^{-1}(\emptyset)=\emptyset$, all are open sets of $X$. – user Jul 05 '16 at 00:51