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.


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?


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.

  • $\begingroup$ Yes, of course! Thank you! I wish it wasnt 13 years since i took my real analysis course :) $\endgroup$ – JKnecht Feb 14 '16 at 9:49

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