# Let $D_k$ be open and dense in $\Bbb K^n$. Prove that $\bigcap D_k$ is dense in $\Bbb K^n$

Im not sure if this proof is correct. In particular it seems that I can use an analogous argument to show that $$\bigcap_{n\in\Bbb N}(0,1/n)\neq\emptyset$$

which is false. The strategy that I used in the proof have it origin in a hint provided by the book were the exercise comes from. In the book it was hinted that we can define $F_k$ and $\bar{\Bbb B}(x_k,r_k)$ as I did below.

But, as I said before, it seems possible that using the same strategy we can prove false things. Maybe I used the hint provided by the book in a bad way, can someone take a look and verify or correct what I did? Thank you.

Let $D_k$ for all $k\in\Bbb N$ be open dense subsets of $\Bbb K^n$ and $D:=\bigcap_k D_k.$ Show that $D$ is dense in $\Bbb K^n$.

By a previous exercise we knows that finite intersections of $D_k$ are open and dense in $\Bbb K^n$. Define $F_k:=\bigcap_{j=0}^k D_j$, then $F_k\supseteq F_{k+1}$ for all $k\in\Bbb N$.

Now choose some $x_0\in\Bbb K^n$ and some $r_0>0$, then we have that $\Bbb B(x_0,r_0)\cap F_0$ is open, then exists some $x_1\in\Bbb K^n$ and some $r_1>0$ such that $\bar{\Bbb B}(x_1,r_1)\subsetneq \Bbb B(x_0,r_0)\cap F_0$, and $\Bbb B(x_1,r_1)\cap F_1$ is also open. Inductively we can define

$$\bar{\Bbb B}(x_{k+1},r_{k+1})\subsetneq\Bbb B(x_k,r_k)\cap F_k,\quad\forall k\in\Bbb N$$

and

$$F_k\supseteq F_{k+1}\implies \bar{\Bbb B}(x_k,r_k)\supsetneq\bar{\Bbb B}(x_{k+1},r_{k+1})$$

hence, by the nested interval property of $\Bbb R$, for every canonical projection

$$\pi_k:\Bbb K^n\to\Bbb K,\quad \langle x^{(0)},x^{(1)},\ldots,x^{(n)}\rangle\mapsto x^{(k)}$$

we knows that the sequence $(x_k^{(j)})$ converges to some $x^{(j)}\in\Bbb K$, hence $(x_k)\to x$ for some $x\in\Bbb K^n$, so $D$ cannot be empty. And because $x_0$ was arbitrary then $D$ is dense in $\Bbb K^n$.$\Box$

Expanding my question: I said that it seems that the same strategy can be used to prove a false statement as

$$\bigcap_{n\in\Bbb N}(0,1/n)\neq\emptyset$$

Let see why I think that: define $F_k=(0,1/k)$ and $\Bbb B(x_k,r_k)$ with $r_k:=\frac12\left(\frac1k+\frac1{k+1}\right)$. Hence we can write an analogous statement as in the proof above

$$\bar{\Bbb B}(x_{k+1},r_{k+1})\subsetneq\Bbb B(x_k,r_k)\cap F_k,\quad\forall k\in\Bbb N$$

• This is the Baire category theorem. Feb 5, 2017 at 18:46
• How do you prove false things from above? How does your first example illustrate this? Feb 5, 2017 at 18:59
• @copper.hat I will edit to show my concerns Feb 5, 2017 at 19:00
• I'm still not sure what you mean. The $F_k$ above is only dense in $[0,{1 \over k}]$. You need $F_k$ to be open & dense in whatever complete space you are looking at. Feb 5, 2017 at 19:22
• You can't. You need to create a collection of nested closed intervals that lie in $F_k$. Given any closed interval $I \subset (0,1)$ there is some $k$ such that $I \cap F_k = \emptyset$. Feb 5, 2017 at 19:37

Working with your dense and open sets $F_k$ that are decreasing, so that $\forall k, F_{k+1}\subset F_k$, Let $x \in X$ and let $C_0$ be a compact neighbourhood of $x$. Then $C_0$ intersects $F_1$, we find a compact subset with non-empty interior $C_1 \subset C_0$ such that $C_1 \subset F_1 \cap C_0$. This intersects $F_2$ again, so we find a compact set $C_2$ with non-empty interior such that $C_2 \subset F_2 \cap C_1$, and we keep going by recursion.
We find $C_k$ compact with $\forall k : C_{k+1} \subset C_k \cap F_{k+1}$
As the $C_K$ are decreasing and compact we find $p \in \cap_n C_n$ which also lies $\cap _n F_n$. As all this sits in $C_0$ we see that $x \in \overline{\cap_n F_n}= \overline{\cap_n D_n }$, as required.