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Let X, Y and Z metric spaces, and $f:X \to Y$ and $g:f(X) \to Z$ continuous functions (where $f(X)$ is a subspace of Y). Need to show that $h(x)=g(f(x))$ is also continuous.

To prove it, I used the proposition that says that, when $f:X \to Y$ is a continuous function between two metric spaces, a sequence $\{x^n\} \to x$, then $f(x^n) \to f(x)$.

So then, $h(x^n)=g(f(x^n)) \to h(x)=g(f(x))$ Thus, $h$ is continuous $\square $.

However, the solutions I have seen on the internet uses the concept that the preimage of open sets are open when $f$ is continuous. These are much longer proofs, so I'm suspecting that my proof is incorrect.

user3889486
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1 Answers1

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Your proof is fine, being able to commute limits of sequences and functions is one of the many definitions of continuity.

The reason you are seeing different proofs online is that your proof requires a different degree of structure on your space. Using the open set definition, you can prove this fact for the most general topological spaces.

P.S Be careful, $x^n$ does not mean the same thing as $x_n$, the latter is the notation for a sequence of $x_n's$

operatorerror
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  • By the way, do you know the most general setting where commuting with limits is equivalent to being continuous? I imagine it has to be smaller than general Hausdorff spaces, even though all nets have unique limits in Hausdorff spaces. – Chill2Macht May 07 '16 at 20:40
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    I am not sure, my instinct was Hausdorff but I had second thoughts (I think you are right that it is smaller) but I don't know enough topology to say – operatorerror May 07 '16 at 20:41