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Continuous map

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Grigorios Kostakos
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Continuous map

#1

Post by Grigorios Kostakos » Thu Oct 13, 2016 1:40 pm

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 $E_1$.
Grigorios Kostakos
jason1996
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Re: Continuous map

#2

Post by jason1996 » Sat Nov 12, 2016 9:08 pm

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.
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Riemann
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Re: Continuous map

#3

Post by Riemann » Tue Nov 29, 2016 2:25 pm

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.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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