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Let be a sequence of pointwise convergent, and continuous functions $f_n : M \to \mathbb {R}$ , where M is a metric space. Prove that $\forall c\in \mathbb R$ , the set $ f^{-1} ([c,\infty ))$ is a $ G_\delta$. Where $ f$ $=$ lim $f_n$

First I know that every $f_n^{-1} ([c,\infty ))$ is closed, since each $f_n$ is continuous. I don't know how to relate that sets and the set $f^{-1} ([c,\infty ))$ . I only noticed one thing and it´s this equality.

$$ f^{ - 1} \left( {\left[ {c,\infty } \right)} \right) \supset \bigcap\limits_{n = 1}^\infty {f_n ^{ - 1} \left( {\left( {c - \frac{1} {n},\infty } \right)} \right)} $$ or instead of $ {\left( {c - \frac{1} {n},\infty } \right)} $ we can consider $ {\left( {\left( {c - h\left( n \right),\infty } \right)} \right)} $ where $h(n)$ it's any positive sequence such that converges to zero. With this I can only conclude that this set contains a $ G_\delta$ but nothing else :(

EDIT: And I also proved that these two sets , not necesarly are equal. For example consider $ f_n \left( x \right) = \frac{x} {n} $ and the interval $[-1,\infty)$

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up vote 1 down vote accepted

Hint: $f(x) \ge c$ iff for every positive integer $n$ there is $m$ such that $m > n$ and $f_m(x) > c - 1/n$.

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I did it thx for your advice –  Matias Jun 3 '12 at 22:59
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