A function $f: X \to Y$ is continuous if and only if $ ^{−1} (C) $ is closed in $X $ for every closed set $C$ in $Y$.

Question

Since a mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous on $X$ if and only if $ ^{−1} (V)$ is open in $X$ for every open set $V$ in $Y$ and since a set is closed if and only if its compliment is open, $^{−1} (E^c)= [^{−1} (E)]^c$ for every $E⊂Y$.

Is this a correct proof of this corollary from Rudin?

Please see https://math.meta.stackexchange.com/questions/5020/ — Angina Seng, Dec 08 '19 at 03:34

score 1 · Answer 1 · answered Dec 08 '19 at 03:43

1

Yes, the key is that the preimage behaves nicely with all the set operations.

answered Dec 08 '19 at 03:43

Martin Argerami

205,756

so nothing needs to change in the proof right? – Mike Langis Dec 08 '19 at 03:49
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No, unless you want to give more detail on the properties of the preimage. – Martin Argerami Dec 08 '19 at 04:05

A function $f: X \to Y$ is continuous if and only if $ ^{−1} (C) $ is closed in $X $ for every closed set $C$ in $Y$.

1 Answers1