Consider $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ . Prove that if $f$ is continuous, then the graph of $f$ is a closed subset of $\mathbb{R}^n \times \mathbb{R}^m = \mathbb{R}^{n+m}$
any ideas on how to commence this problem?
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2Note: A more general variant of this questions seems to be Is the graph Gf .. a closed subset of X×Y?. – mvw Nov 04 '15 at 00:59
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Can you at least start? Say take a sequence of points in the graph that converges to a point in $\mathbb R^n\times \mathbb R^m?$ – zhw. Nov 04 '15 at 00:59
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See my answer to this question, http://math.stackexchange.com/questions/1505745/is-a-vector-field-a-closed-map/1505755#1505755. The same proof works here. – jgon Nov 04 '15 at 01:04
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Brian's answer seems a reasonable approach. He goes for showing that the complement of the graph is open. – mvw Nov 04 '15 at 01:05
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Take $(x_n,f(x_n))\in G(f)$ auch that $(x_k,f(x_k))\to (x,y)$. So you have $x_k\to x$ in $\mathbb{R}^n$ and by continuity of $f$ you have $y=\lim_kf(x_k)=f(x)$. Therefore $(x,y)=(x,f(x))\in G(f)$.
Euler88 ...
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Consider the projections $\pi_1$ and $\pi_2$.
The graph of $f$ is
$$(\pi_2-f\circ \pi_1)^{-1}(0),$$
hence closed.
Aloizio Macedo
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