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$(X,d)$ be a complete metric space,$f:X\to\mathbb{R}$ be such that $\forall\alpha\in\mathbb{R},\{x:f(x)<\alpha\}$ is an open set in $X$. I need to prove that set of points of continuities of $f$ is dense in $X$

I know these stuffs which I intuitively think I need to apply to solve the problem but not able to bring out any fruitful complete logic.

  1. Set of points of continuity of a function can be written as countable intersection of open sets. ($G_{\delta}$ set)

  2. A complete metric space can not be written as a countable union of nowhere dense sets or if $A_n,n\in\mathbb{N}$ be open dense set such that $\cap A_n\ne \emptyset$ then $\cap A_n$ is dense.

  3. In a complete metric space every cauchy sequence has its limits in it.

Could anyone help me to start solving the problem ?

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  • $\begingroup$ If you write the continuity set of $f$ as the intersection of countably many open sets, can you prove that all these open sets are dense? $\endgroup$ Oct 6 '13 at 13:45
  • $\begingroup$ How do you define the $A_n$? To show a set is dense, you need some form of description of it. $\endgroup$ Oct 6 '13 at 13:57
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This can be done in 2 steps :

  1. For $a < b$ in $\mathbb{R}$ define $$ A_{[a,b]} = \{x \in X : f(x) \geq b, \text{ and } \exists (x_j), x_j\to x, \text{ such that } \lim_j f(x_j) \leq a\} $$ Show that $A_{[a,b]}$ is closed and has empty interior.
  2. For each $k\in \mathbb{Z}, m\in \mathbb{N}$, define $$ B_{k,m} = A_{\left[ \frac{k-1}{m}, \frac{k}{m} \right]}^c $$ Show that, if $$ x \in \bigcap_{k,m} B_{k,m} $$ then $f$ is continuous at $x$.

Now you can apply Baire category.

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  • $\begingroup$ It would be good if you give me full solution, I am not able to do $\endgroup$
    – Marso
    Oct 7 '13 at 5:37
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It seems the following.

Assume the converse. Let $U\subset X$ be a non-empty open set without continuity points of the map $f$ in it. Then for each point $x\in U$ there exist a number $n(x)$ and a sequence $\{x_i\}$ of points of $X$ converging to the point $x$ such that $$|f(x_i)-f(x)|\ge 1/n(x)$$ for each index $i$.

Since a set $$\{y\in X : f(y)<f(x)+1/(2n(x))\}$$ is open, refining the sequence $\{x_i\}$ to a subsequence, if necessarily, we may assume that $$f(x_i)-f(x)\ge 1/n(x)$$ for each index $i$.

For each natural $n$ put $$U_n=\{x\in u:n(x)=n\}.$$ Since the set $U$ is a non-empty open subset of a complete metric space, it is Baire, so there exist a non-empty open subset $V$ of the set $U$ and a number $n$ such that $U_n$ is dense in $V$. Let $x\in V\cap U_n$ be an arbitrary point. There exists a sequence $\{x_i\}$ of points of $X$ converging to the point $x$ such that $$f(x_i)-f(x)\ge 1/n(x)$$ for each index $i$.

Since a set $$W=\{y\in X: f(y)<f(x)+1/(2n)\}$$ is open, there exists a point $x_i\in W$, a contradiction.

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