Let $f: (X,d) \to (Y,d')$ an open function (not necessarily continuous) between metric spaces. Decide whether the following statements are true or false:

1) If $A \subset X$ doesn't have isolated points, then neither does $f(A)$.

2) If $B \subset Y$ doesn't have isolated points, then neither does $f^{-1}(B)$.

My attempt at a solution:

I think that 1) is a false statement, so I am trying to find a counterexample. Is the following correct?:

Let $(X,d)$ and $(Y,d')$ be metric spaces such that $X=\mathbb R$, $\space$ $d'$ is the discrete metric and $|Y|\geq 2$. Now let $a \in \mathbb R$ and consider the ball $B(a,\delta)$ for some $\delta>0$. Given $y_1, y_2 \in Y$ with $y_1 \neq y_2$ If we define $f$ such that $f(x)=y_1$ if $x \in B(x,\delta)$ and $f(x)=y_2$ if $x \not \in B(a,\delta)$, then clearly $f$ is an open function and $B(a,\delta)$ has no isolated points, however, $f(a)=y_1$ is an isolated point in $Y$.

For 2), maybe I could prove the contrapositive: if $f^{-1}(B)$ has isolated points, then $B \subset Y$ has isolated points. I don't see how to prove this, I would appreciate some help for 2).


Your answer to (1) seems absolutely fine; you are taking advantage of the fact that any mapping into a discrete space is (not necessarily continuous but) open, and every point in a discrete space is isolated. A somewhat simpler (to my eyes, at least) example can be had as follows: denote by $\mathbb{R}_{\text{d}}$ the reals with the discrete metric, and now consider the identity map $\mathrm{id}_\mathbb{R} : \mathbb{R} \to \mathbb{R}_{\text{d}}$, and any $A \subseteq \mathbb{R}$ without isolated points, such as $\mathbb{R}$ itself.

For (2) I'll just provide a

Hint: If $x$ is an isolated point of $f^{-1} [ B ]$, then there is a $\varepsilon > 0$ such that $B ( x; \varepsilon ) \cap f^{-1} [ B ] = \{ x \}$. Note that $f [ B ( x ; \varepsilon ) ] \subseteq Y$ is open. What happens if $y \in f [ B ( x ; \varepsilon ) ] \cap B$ is distinct from $f(x)$? (In particular, what does "$y \in f [ B ( x ; \varepsilon ) ]$" mean?)

  • $\begingroup$ Got it. If there is $y \in f[B(x,\epsilon)] \cap B$ : $y \neq f(x)$, then, there is $x'\in B(x,\epsilon)$ : $f(x')=y$, since $y \neq f(x) \implies x' \neq x$. But then, $x' \in B(x,\epsilon) \cap f^{-1}[b]=\{x\}$, which is clearly absurd. We've found a set $S \subset Y$ such that $S$ has isolated points. We've proved the contrapositive holds, which means the statements is true. $\endgroup$ – user100106 Dec 7 '13 at 6:42
  • $\begingroup$ BTW, for 1) your example is simpler, but I am just happy to know mine is correct. Thanks very much! $\endgroup$ – user100106 Dec 7 '13 at 6:43
  • $\begingroup$ @user100106: Yup, you've got it! (And you're welcome.) $\endgroup$ – user642796 Dec 7 '13 at 6:44

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