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Math Help - Sequentially Compact and composition

  1. #1
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    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)=<br />
\begin{cases}<br />
1 & \text{if $x \in Q$}, \\<br />
0 & \text{if $x \notin Q$}.<br />
\end{cases}<br />
    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.
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    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.
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  3. #3
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    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.
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