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${f_n}$ is a sequence of (real-valued) functions uniformly converging to $f$.

Suppose further that $f_n(x)< M$ (The functions $f_n$ are uniformly bounded above by $M$).

Is it true that $f(x) < M$ everywhere?

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    $\begingroup$ $f(x) = 1, f_n(x) = 1 - 1/n, M = 1$. Answer: no. $\endgroup$ – Savio Dec 2 '15 at 20:11
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    $\begingroup$ +1 but what about $\leq$... $\endgroup$ – parsiad Dec 2 '15 at 20:12
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    $\begingroup$ $\le$ yes, because if $f(x) > M$ then $f_n(x) > M$ for $n$ sufficiently large, contradiction $\endgroup$ – Savio Dec 2 '15 at 20:14
  • $\begingroup$ +1 You should consider posting these two comments as an answer :) $\endgroup$ – parsiad Dec 2 '15 at 20:14
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Let $f(x)=1$, $f_n(x)=1−1/n$ and $M=1$. Answer: no.

But it is true if $<$ is replaced by $\le$, because if $f(x)>M$ then $f_n(x)>M$ for $n$ sufficiently large, contradiction.

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