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### Continuous map

Posted: **Thu Oct 13, 2016 1:40 pm**

by **Grigorios Kostakos**

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 $E_1$.

### Re: Continuous map

Posted: **Sat Nov 12, 2016 9:08 pm**

by **jason1996**

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.

### Re: Continuous map

Posted: **Tue Nov 29, 2016 2:25 pm**

by **Riemann**

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}

= g_1^{-1}\cup g_2^{-1}$$ 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.