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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Possible duplicate of Equivalence of continuity definitions $\endgroup$ – user99914 Dec 27 '17 at 13:51

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.

  • $\begingroup$ 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

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