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Suppose that $X$ is a space with the property that for any point $p \in X$ there is a map $f: X \rightarrow \mathbb{R}$ such that $f^{-1}(1) = \{p\}$. Show that $X$ is Hausdorff.

Solution:

Suppose that $x \neq y$ are points in $X$. Choose $f: X \rightarrow \mathbb{R}$ with $f(x) = 1$ and $f(y) \neq 1$ and open disjoint intervals around $1$ and $f(y)$ in $\mathbb{R}$ (which is possible since $\mathbb{R}$ is Hausdorff).

Now we are done if we can say that the inverse images of these intervals under $f$ are open disjoint sets in $X$ containing $x$ and $y$, respectively. Because then $X$ is Hausdorff.

$f$ is a map, i.e. a continuous function, and therefore its inverse takes open sets to open sets.

This is probably a very simple question and bordering on stupid for you experienced topologists but i just wonder how we can be sure that the inverse takes disjoint open sets to disjoint open sets?

Martin Sleziak
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JKnecht
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  • On preimage of intersection: http://math.stackexchange.com/questions/228711/what-are-the-strategies-i-can-use-to-prove-f-1s-cap-t-f-1s-cap-f or http://math.stackexchange.com/questions/516374/how-to-prove-f-1b-1-cap-b-2-f-1b-1-cap-f-1b-2 or http://math.stackexchange.com/questions/144870/verifying-a-proposition-on-image-and-preimage-fa-cap-b-subseteq-fa-cap-fb – Martin Sleziak Apr 22 '16 at 05:46
  • The condition on $X$ can also be expressed by saying that all singletons are zero sets. – Martin Sleziak Apr 22 '16 at 05:55

1 Answers1

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It's true for every map that the inverse image of disjoint sets are disjoint sets.

Let $f\colon X\to Y$ be any map between sets, $A, B\subseteq Y$, $A\cap B=\emptyset$. Suppose there is an element $x\in f^{-1}(A)\cap f^{-1}(B)$. Then we have $f(x)\in A\cap B=\emptyset$ a contradiction.