0
$\begingroup$

Let $X$ a topological space and $(Y, d)$ a metric space. Let $f: X \rightarrow Y$ a function. Then, $f$ is continuous iff $\forall \, x \in X \, \, \, \forall \, \epsilon >0 \, \, \, \exists \, U \subset X$ open neighborhood of $x$ such that $d(f(x), f(y)) < \epsilon \, \, \forall \, y \in U$.

I know how to prove a similar statement for a map between metric spaces (using as definition that if $f$ is continuous, then $f^{-1}(U)$ is open for all $U \subset Y$ open set), but it didn't worked in this case.

Any suggestions? Thanks in advance!

$\endgroup$
  • $\begingroup$ A metric space has a basis of open balls. A function is continuous iff the pre-image of a basis element is an open set. This last is easily proven: look up basis and open ball. $\endgroup$ – Tom Collinge Mar 2 '18 at 14:08
  • $\begingroup$ I tried to use your hint. If $\mathbb{B}$ is basis of Y (so the set of all open balls in $Y$) and $B_\alpha \in \mathbb{B}, \alpha \in J$, then we have by definition that $U\subset Y$ open iff $\exists \, B_\alpha \in \mathbb{B}, \alpha \in J$, such that $U = \cup_{\alpha \in J} B_\alpha$. So $f$ is continuous iff $ f^{-1}(U)= f^{-1}(\cup_{\alpha \in J} B_\alpha) = \cup_{\alpha \in J} \, f^{-1}(B_\alpha)$ is open. But how can I say that $f^{-1}(B_\alpha)$ is open? $\endgroup$ – userr777 Mar 2 '18 at 18:42
  • $\begingroup$ Sorry wrong question, I meant how can I connect this fact with the existence of the neighborhood $U$ such that $f(U) \in B_{\epsilon}(f(x))$? $\endgroup$ – userr777 Mar 2 '18 at 20:40
  • $\begingroup$ There a quite a few conditions that are equivalent to continuity of $f:X\to Y,$ for any spaces $X,Y,$ including this one, if you replace the set open $d$-balls centered $f(x)$ with the set of either (i) every open $V \subset Y$ such that $f(x)\in V,$ or (ii) every $W$ such that $f(x)\in W\in \Bbb B,$ where $\Bbb B$ is a base (basis) for $Y$ ...Some problems & constructions seem to be more suited to one condition than to another. $\endgroup$ – DanielWainfleet Mar 3 '18 at 15:00
3
$\begingroup$

From the condition to continuity of $f$: let $O$ be open in $Y$. We want to show that $f^{-1}[O]$ is open in $X$. So let $X \in f^{-1}[O]$, so we know that $y:=f(x) \in O$. As $O$ is open in the metric space $(Y,d)$ we have some $\varepsilon>0$ such that $$B(y, \varepsilon) \subseteq O$$

Applying the condition to $x,\varepsilon$ we get some open neighbourhood $U$ of $x$ such that for all $p \in U$ we have $d(y, f(p)) < \varepsilon)$. This means that for all $p \in U$, $f(p) \in B(y,\varepsilon)$ and so for all $p \in U$, $f(p) \in O$ and so $$\forall p \in U: p \in f^{-1}[O]$$

and this shows that $U \subseteq f^{-1}[O]$ and so $x$ is an interior point of $f^{-1}[O]$. As $x$ was arbitary, $f^{-1}[O]$ is open in $X$.

From continuity to the condition is similar and even simpler: given $x$ and $\varepsilon >0$ the set $O= B(f(x), \varepsilon)$ is open in $Y$ so $x \in f^{-1}[O]$ and the latter set is open (as $f$ is continuous), call it $U$ and check it obeys what we want: all points $y$ in $U$ map to $O$ which is the open ball so $d(f(y), f(x)) < \varepsilon$.

$\endgroup$

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.