Sequentially Compact and composition

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October 16th 2008, 06:31 PM
Paperwings

Sequentially Compact and composition

I had an exam question today along the lines of:

Prove or disprove: If $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the graph of f defined as $G = \left\{ (x,f(x)) | x \in \mathbb{R} \right\}$ is sequentially compact(something else was also sequentially compact, which I believe f was), then f must be continuous.

I do not know if this is true or false. Quite honestly, I can not picture this in my head. By counterexample of a characteristic function:
$f(x)= \begin{cases} 1 & \text{if $x \in Q$}, \\ 0 & \text{if $x \notin Q$}. \end{cases} $
I would think to be false, but I don't think this can be rewritten as the composition of two functions.

Conversely, if f was continuous, then for h(x) = g(x,f(x)) would be continuous since it is the composition of continuous functions.

Thank you for reading. Any help is greatly appreciated.
October 16th 2008, 06:37 PM
ThePerfectHacker

Quote:

Originally Posted by Paperwings

I had an exam question today along the lines of:

Prove or disprove: If $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the graph of f defined as $G = \left\{ (x,f(x)) | x \in \mathbb{R} \right\}$ is sequentially compact(something else was also sequentially compact, which I believe f was), then f must be continuous.

The question does not make sense. The set $G = \{ (x,f(x)\}$ is never sequentially compact. Because we can choose the sequence $\bold{x}_n = (n,f(n)) \in G$ . This sequence is unbounded since the first component is unbounded. Therefore, $\bold{x}_n$ has no convergent subsequence.
October 16th 2008, 06:54 PM
Paperwings

I think I worded it incorrectly. I know it was an "if... then" statement where the then statement was "f is continuous". Oh well, thanks TPH.