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Given metric spaces $B$ and $P$, a function $q: B \to P$ is continuous at $c \in B$ if for every $\epsilon > 0$, there exists $\delta > 0$ such that

$$d_B(x, c) < \delta \implies d_P(q(x), q(c)) < \epsilon$$

But if $B$ and $P$ happen to be topological spaces, $q$ is continuous if the preimage of every open subset of $P$ is open in $B$. So in this case, what would it mean for $q$ to be continuous at $c \in B$?

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Related question:… – Jonas Meyer Jul 1 '12 at 10:29

1 Answer 1

$q$ is continuous at $c\in B$ if and only if for every neighborhood $W$ of $q(c)$ there exists a neighborhood $U$ of $c$ such that $q(U)\subseteq W$. You may replace "neighborhood" with "open set that contains".

If you translate what this means in the case of the topology induced by a metric, you will find that it is exactly the usual $\epsilon$-$\delta$ definition.

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An interesting fact is that the "exists a neighborhood $U$ such that $q(U) \subset W$" condition can be replaced by "$f^{-1}(W)$ is a neighborhood of $c$". I had a similar question here:… – Eric Auld May 24 '14 at 1:13
A neighborhood can be non-open. By "$U$ is a neighborhood of point $c$", we mean "there is an open set $M$ such that $c \in M \subseteq U$", but $U$ itself can be non-open. – jdh8 Sep 27 at 10:27

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