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I have a (probably stupid) question about the Baire Category Theorem. I am looking at the statement that says that in a complete metric space, the intersection of countable many dense open sets is dense in the metric space. Assume that we have the countable collection of dense open sets $\{U_n\}$ in a complete metric space $X$, and let $x \in X, \epsilon>0$. Since $U_1$ is dense in $X$, there is $y_1\in U_1$ with $d(x,y_1)<\epsilon$. Also, as $U_1$ is open, there is $r_1>0$ with $B(y_1;r_1)\subset U_1$. Then, we can arrange $r_1<1$ such that $\overline{B(y_1;r_1)} \subset U_1\cap B(x;\epsilon)$. Now my question is why we can arrange that the closure will be contained in each of them? I think intuitively it sounds correct, but I didn't succeed to prove it rigorously. Can you please help me here?

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  • $\begingroup$ The correct statement is that the intersection of countably many dense open subsets is dense, not open. Eg the irrational numbers are not open in the reals but are of that form. $\endgroup$ – Henno Brandsma Jun 10 '18 at 21:05
  • $\begingroup$ Follow-up to the above comment. It seems that you made a typo, saying "open" when you meant to say"dense". If so, it would be better if you edited your Q before you get any negative comments. $\endgroup$ – DanielWainfleet Jun 10 '18 at 21:47
  • $\begingroup$ Since the intersection of a non-empty finite family of dense open sets is also dense and open, the proof is a little easier if we begin by letting $V_n=\cap_{j\leq n}U_j,$ since $V_{n+1}\subset V_n.$ We have $\cap_{n\in \Bbb N}V_n=\cap_{n\in \Bbb N}U_n,$ and each $V_n$ is dense & open. $\endgroup$ – DanielWainfleet Jun 10 '18 at 21:52
  • $\begingroup$ Yes, I've just noticed it. Corrected it. Thank you. $\endgroup$ – Mr. Tea Jun 11 '18 at 20:39
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$d(x, y_1) < \varepsilon$ actually means $y_1 \in B(x; \varepsilon)$. Also $y_1 \in U_1$ by construction.

Hence $y_1 \in U_1 \cap B(x;\varepsilon)$. Now, since $U_1 \cap B(x;\varepsilon)$ is an open neighbourhood of $y_1$, there exists $r < 1$ such that $B(y_1, r) \subseteq U_1 \cap B(x;\varepsilon)$. Now define $r_1 = \frac{r}2$ and notice

$$\overline{B(y_1, r_1)} = \overline{B\left(y_1, \frac{r}2\right)} \subseteq B(y_1, r) \subseteq U_1 \cap B(x;\varepsilon)$$

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