2
$\begingroup$

I was trying to prove the equivalence of the $\epsilon$-$\delta$ and topological notions of continuity at a point. (Given the standard topology on a metric space) I could get one direction, but the $\epsilon$-$\delta$ notion of continuity at a point doesn't seem to imply the topological notion of continuity at a point. (I think I might have messed up my definition)

The definition of topological continuity at a point I was using was that a function $f$ is continuous as point $a$ if every open set in the image of $f$ which contain $f(a)$ has an open pre-image.

Basically, the $\epsilon$-$\delta$ notion of continuity at point $a$ only says things about neighborhoods of $a$. But I can always union a neighborhood of $a$ with an open set in some other part of the image to get a new open set. And the $\epsilon$-$\delta$ definition gives me no information about this potentially distant set or its pre-image.

In other words, take $f$ to map some open set $A$ to some open neighborhood $f(A)$ and some closed set $B$ to some open neighborhood $f(B)$ such that $f(A) \cap f(B)=\emptyset$. Also, let it be that f is $\epsilon$-$\delta$ continuous over all of $A$.

So now, let's take some $a \in A$. $f$ is epsilon-delta continuous at $a$. But is it topologically-continuous at $a$? No. Because any open neighborhood of $f(a)$, I can union with $f(B)$ to get an open set whose pre-image is not an open set.

I think my problem is that I got my topological definition of continuity at a point wrong. But I can't figure out how to fix it without invoking concepts from metric spaces.

$\endgroup$
  • 1
    $\begingroup$ a function $f$ is continuous at point $a$ if every open set in the image of $f$ which contain $f(a)$ has an open pre-image - I think this is a stronger condition than continuity at $a$. Continuity at $a$ only requires that every neighbourhood of $f(a)$ has some corresponding open neighbourhood of $a$ which maps inside of it. $\endgroup$ – Myridium Dec 27 '17 at 0:33
  • $\begingroup$ I appreciate the pointer, but the question is different because I was looking for continuity at a point specifically. The question linked to speaks uses the "the pre-image of open sets is open" definition which is a stricter notion. $\endgroup$ – azani Dec 27 '17 at 0:42
  • $\begingroup$ Yep you're right, sorry. There is also this similar question $\endgroup$ – Myridium Dec 27 '17 at 1:18
  • 2
    $\begingroup$ no! The notion of neighbourhood does not require the notion of a metric: A neighbourhood $V$ of a point $u$ is a set containing an open subset $a \in U \subseteq V$. $\endgroup$ – Myridium Dec 27 '17 at 1:30
  • 1
    $\begingroup$ Possible duplicate of Equivalence of continuity definitions $\endgroup$ – user99914 Dec 27 '17 at 13:51
3
$\begingroup$

According to M. Winter's answer here, we can define $f$ to be continuous at $a$ if every open set in the image of $f$ which contains $f(a)$ has a pre-image which contains an open set containing $a$.

In your example, the union $A \cup B$ (the pre-image of $f(A) \cup f(B)$) isn't open, but it does contain an open set $A$ containing $a$, so this definition will work as you intend.

$\endgroup$
  • $\begingroup$ To the proposer: Observe that for $f:\Bbb R\to \Bbb R$ with the usual topology on $\Bbb R,$ the def'n of continuity of $f$ at $a\in \Bbb R$ in this answer is the same as the classical $\epsilon$-$\delta$ def'n. $\endgroup$ – DanielWainfleet Dec 27 '17 at 17:19

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.