Path-Connectedness in Uncountable Finite Complement Space

Question

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.

Answer 1

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.

Answer 2

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.