# Open set/continuous functions

$(X,d_X)$ is a metric space and $x \in X$. How to prove:

{$x$} $\subset X$ is an open set $⇔ f: X \to \mathbb{R}$ is continuous in $x$ for all $f$

I tried for $\Rightarrow$

Since {$x$} $\subset X$ is an open set:

$\forall$ $x \in$ {$x$} $\exists$ $\varepsilon > 0: B_\varepsilon (x) \subset$ {$x$}.

Since $x$ is an isolated point

$\forall \varepsilon > 0 \exists \delta >0$: $|x-x_0|<\delta \Rightarrow x=x_0 \Rightarrow |f(x)-f(x_0)|=0<\varepsilon$.

I chose $\delta = \varepsilon$. So it has to be continuous.

For $\Leftarrow$ I don't know how to conclude from the continuity to the open set.

• HINT: if every singleton set is open in $X$, then EVERY subset of $X$ is open. Use the "preimage of an open set is open" definition of continuity. – Joe May 21 '18 at 13:37

Let $d$ be the discrete topology on $X$. Then the identity function $\operatorname{id}$ from $(X,d_X)$ into $(X,d)$ is continuous at $x$. Since $\{x\}$ is an open subset of $(X,d)$, $\operatorname{id}^{-1}(\{x\})$ is an open subset of $(X,d_X)$; in other words, $\{x\}$ is an open subset of $(X,d_X)$.