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Steen & Seebach, Counterexamples in Topology, have the following as problem 30 on p. 206:

Every countable $T_1$ space is totally path disconnected.

I can see why a Hausdorff countable space $X$ is totally path disconnected, i.e., all paths into $X$ are constant. Suppose there is a non-constant path $f:[0,1]\to X$ from $x$ to $y$ with $x\ne y$. The image of the path is then path-connected and Hausdorff. And such a space is arc-connected (Willard, Corollary 31.6, consequence of the Hahn-Mazurkiewicz theorem). Since $x$ and $y$ are distinct points, any arc joining them has the cardinality of the continuum, and so the space $X$ cannot be countable.

What is the correct argument if we just assume $T_1$?

PatrickR
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1 Answers1

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Suppose you had a continuous non-constant $f:[0,1]\to X$.

Since $X$ is $T_1$, the one-element set containing $p$ is a closed set for all $p\in X$.

Therefore $[0,1]$ would be the union of countably many (and more than one, because the function is not constant) disjoint nonempty closed sets $f^{-1}(p)$.

But this is impossible: Is $[0,1]$ a countable disjoint union of closed sets?

PatrickR
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Chris Sanders
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  • Nice. And the image of $f$ (non-constant) cannot be finite either, as that would provide a nontrivial finite partition of $[0,1]$ by closed sets, which is not possible because $[0,1]$ is connected. – PatrickR Jun 19 '22 at 20:13
  • @PatrickR . Countable means "finite or countably infinite" so that uncountable means the same as not countable. – DanielWainfleet Jun 20 '22 at 01:49