If $f\vert_{K}\in\mathcal{C}(K,N)$, then $f\in\mathcal{C}(M,N)$.

Question

Let $f:M\to N$, show that if $K\subset M$, $K$ is a compact subset of $M$, and if $f\vert_{K}\in\mathcal{C}(K,N)$, then $f\in\mathcal{C}(M,N)$.

My approach: If $f\vert_{K}$ on the space of continuous functions $\mathcal{C}(K,N)$, such that $f\vert_{K}:K\to N$, with $f\vert_{K}=f\circ id$, where $id:K\to M$ and $id(x)=x$ for all $x\in K$. Then, if $f\vert_{K}$ is continuos, also is $f\circ id$, therefore $f$ is continuos. This is right?? regards!

Is the statement perhaps supposed to hold for all compact subsets $K$? As noted by Paul Sinclair, it's false as written, and your proof doesn't work: knowing that $f \circ id$ is continuous doesn't let you conclude that $f$ is. — Nate Eldredge, Oct 08 '15 at 23:52

score 1 · Accepted Answer · answered Oct 08 '15 at 23:46

This is wrong. The statement is clearly false. For example, Let $M = N = \Bbb R$, and let $$f(x) =\begin{cases} 1 & x \in [0, 1] \\ 0 & x \notin [0,1]\end{cases}.$$ Then $f$ is continuous on $[0,1]$, but is not continuous on $\Bbb R$.

More generally note that there is absolutely nothing given that restricts the function outside of $K$. You can give $f$ any values you like outside of $K$, and as long as it is continuous inside $K$, it satisfies your conditions. With no control outside of $K$, there is no need for the function to be continuous there.

If $f\vert_{K}\in\mathcal{C}(K,N)$, then $f\in\mathcal{C}(M,N)$.

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