1
$\begingroup$

$(X,d_X)$ 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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Joe May 21 '18 at 13:37
0
$\begingroup$

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 $(X,d_X)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.