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  1. #1
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    Real Analysis Help

    Let X be a topological space and let {f_n} be a sequence of continuous functions from X to the real numbers. If f(x)= \lim_{n->\infty} f_n(x) for all x in X, show that the set of points where f is continuous is a G_\delta set.
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  2. #2
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    Hint

    Sorry, I have more info I left off.

    Hint: Let V_n(\epsilon)={x \in X : |f_n(x)-f(x)|<\epsilon} and define O(\epsilon)=\bigcup_{n=1}^\infty Int(V_n(\epsilon)). Then f is continuous on the G_\delta given by \bigcap_{k=1}^\infty O(1/k)
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  3. #3
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    Let U_n = \{x\in X : \exists\text{ op{}en }V_x\ni x \text{ such that }v,w\in V_x\,\Rightarrow\,|f(v)-f(w)|<1/n\}.

    If x\in U_n then V_x\subseteq U_n, so U_n is open. Also, f is continuous at x if and only if x\in\bigcap_nU_n. Thus the set of points of continuity of f is a G_\delta.

    What bothers me about that argument is that it seems to apply to an arbitrary function f. (It doesn't assume that f is the pointwise limit of a sequence of continuous functions.) Am I missing something?
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  4. #4
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    Quote Originally Posted by Opalg View Post
    Let U_n = \{x\in X : \exists\text{ op{}en }V_x\ni x \text{ such that }v,w\in V_x\,\Rightarrow\,|f(v)-f(w)|<1/n\}.

    If x\in U_n then V_x\subseteq U_n, so U_n is open. Also, f is continuous at x if and only if x\in\bigcap_nU_n. Thus the set of points of continuity of f is a G_\delta.

    What bothers me about that argument is that it seems to apply to an arbitrary function f. (It doesn't assume that f is the pointwise limit of a sequence of continuous functions.) Am I missing something?
    I think you're right.

    But this problem (and the suggested solution) also reminds me of something else, which is somewhat more interesting: under the additional hypothesis that X is a complete metric space (or any Baire space), f is continuous on a dense G_\delta set.

    This can be proved by considering V_n(\varepsilon)=\{x\in X|\forall m\geq n, \forall p\geq n, |f_m(x)-f_p(x)|\leq \varepsilon\}. This is a closed subset of X. Then let O(\varepsilon)=\bigcup_{n=1}^\infty {\rm Int}(V_n(\varepsilon)). This is an open subset, and it is dense because of Baire's theorem: if not, then there would be x\in X and a closed ball B=\overline{B}(x,\delta) such that O(\varepsilon)\cap B=\emptyset, which means {\rm Int}(V_n(\varepsilon))\cap B=\emptyset for all n, hence {\rm Int}(V_n(\varepsilon)\cap B)=\emptyset, and Baire's theorem would say {\rm Int}\left(\bigcup_{n=1}^\infty (V_n(\varepsilon)\cap B)\right)=\emptyset, in contradiction with \bigcup_{n=1}^\infty (V_n(\varepsilon)\cap B)=B (which holds because the pointwise convergence gives \bigcup_n V_n(\varepsilon)=X). Finally, Baire's theorem shows that O=\bigcap_{k=1}^\infty O(1/k) is dense. And one can see that f is continuous on O.
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