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Thread: Real Analysis Help

  1. #1
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    Real Analysis Help

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

    Sorry, I have more info I left off.

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

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

    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 $\displaystyle U_n = \{x\in X : \exists\text{ op{}en }V_x\ni x \text{ such that }v,w\in V_x\,\Rightarrow\,|f(v)-f(w)|<1/n\}.$

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

    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 $\displaystyle X$ is a complete metric space (or any Baire space), $\displaystyle f$ is continuous on a dense $\displaystyle G_\delta$ set.

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