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 Author: Grigorios Kostakos [ Thu Oct 13, 2016 1:40 pm ] Post subject: Continuous map Let $E_1\,,\; E_2$ be metric spaces and $f:E_1\longrightarrow E_2$ a map such that for every compact subspace $S$ of $E_1$ the restriction $f|_S:S\subseteq E_1\longrightarrow E_2$ is continuous. Prove that $f$ is continuous on $E_1$.

 Author: jason1996 [ Sat Nov 12, 2016 9:08 pm ] Post subject: Re: Continuous map We suppose that $x_{n}\rightarrow x_{0}$. The set $\left \{ x_{1},x_{2},.... \right.\left. \right \}\cup \left \{ x_{0} \right.\left.\right \}$ is compact and from the hypothesis(the restriction of f in every compact set is continuous) we obtain that $f(x_{n})\rightarrow f(x_{0})$. So the proof has finished.Question: Is this true for topological (not necessarily metric) spaces? If yes prove it, if not find a counterexample.

 Author: Riemann [ Tue Nov 29, 2016 2:25 pm ] Post subject: Re: Continuous map jason1996 wrote:Question: Is this true for topological (not necessarily metric) spaces? If yes prove it, if not find a counterexample.Hello jason1996,well the answer to your question is yes under one condition. $f$ must be defined on either closed or open subsets of the topological space. Try looking up for the final topology.A little bit general result is the following:Let $X$ be a topological space and $A_1$ and $A_2$ be subspaces of $X$. Let $g_{i}:A_i\rightarrow Y$ be continous for $i=1,2$. If $g_1$ and $g_2$ coincide on $A_1\cap A_2$ and $A_1$ and $A_2$ are either both open or both closed, then the function $f:X\rightarrow Y$ defined by $f(x)=g_1(x)$ if $x \in A_1$ and $f(x)=g_2(x)$ otherwise is continuous.Assume that $g_1$ and $g_2$ are given continuous functions that satisfy the hypothesis and $A_1$ and $A_2$ are both open (respectively closed). Let $U$ be an open set (respectively closed set) of $Y$. Then, $$f^{-1}[U] = g_1^{-1}[U]\cup g_2^{-1}[U]$$ which is a binary union of open (respectively closed) sets in $X$, hence open (closed) by continuity of $g_i$ ($i=1,2$). Hence $f$ is continuous.

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