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Let $\{G_n\}$ be a sequence of dense open subsets of a complete metric space $X$. Show that $ \bigcap_{n \geq 1}G_n$ is nonempty.

Proof:

Consider some point $p_1 \in G_1$. There exist real numbers $0< r_1 <r<1$ such that $$N_1 = N_{r_1} (p_1) \subset N_r (p_1) \subset G_1.$$

Let $q$ be a limit point of $N_1$. For any $\epsilon > 0$ we can find $x \in X$ such that $x \in N_1$ and $x \in N_\epsilon (q)$. Hence $$d(p_1, q) \leq d(p_1, x) + d(x, q)<r_1 + \epsilon.$$ Since $\epsilon$ was arbitrary it follows that $$d(p_1, q) \leq r_1.$$

Thus, if $q$ is any limit point of $N_1$, $q$ is an interior point of $G_1$ (to see this, simply consider the neighbourhood of radius $0<\epsilon<r-r_1$) and so we can write

$\overline{N_1} \subset G_1.$

Since the $G_n$ are dense, either $p_1$ is a point in $G_2$ or a limit point of $G_2$. In either case, arguing as above, we can find a point $p_2 \in G_2 $ and a neighbourhood of $p_2$ with radius $r_2 < \min\big(r_1, \frac{1}{2}\big)$ such that $$\overline{N_2} \subset \overline{N_1}$$ and, since the $G_n$ are all open, we also choose $r_2$ small enough that $$\overline{N_2} \subset G_2.$$

Continuing this process, we can construct the sequence of closed subsets $\Big\{\overline{N_n}\Big\}$. Further, by the way we have chosen the $r_n$ (i.e., $0<r_n<\frac{1}{n}$), it follows $r_n \to 0$, so we have actually constructed a nested sequence of closed bounded sets $\Big\{\overline{N_n}\Big\}$ such that $diam \overline{N_n} \to 0.$ Since $X$ is a complete metric space, the intersection of this sequence is nonempty and since each $$\overline{N_n} \subset G_n$$ the result follows. $\qquad \square$

Assuming this proof is correct, I am wondering if it can be extended to show that the intersection of the $G_n$ is actually dense, or if that requires a completely different line of argument?

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  • $\begingroup$ If $X$ is the empty space then each $G_n$ is empty and so is $\cap_nG_n.$ This is a trivial case but sometimes an overlooked case can bite you . $\endgroup$ – DanielWainfleet Aug 23 '18 at 4:39
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That the intersection of the $G_n$ is dense, is a small modification of the above argument: let $O$ be any non-empty open set in $X$, and start with an open ball $N_0 \subseteq O$ and stay inside $N_0$ with all subsequent steps. This hardly takes any effort at all, but does show that the $x \in \cap G_n$ is also in $N_0$ hence in $O$. So the intersection of the $G_n$ intersects every non-empty open set, hence is dense.

The construction of the $N_n$ can be a bit simplified:

  • Start with $p_0 \in O$ and $N_0 := B(p, r_0) \subseteq O$.
  • $N_0$ is open, so $G_1 \cap N_0$ is non-empty (as $G_1$ is dense) and open (as $G_1$ is open). So pick $p_1 \in G_1 \cap N_0$ and let $0< r_1 < 1$ be small enough that $\overline{B(p_1, r_1)} \subseteq G_1 \cap N_0$. Define $N_1 = B(p_1, r_1)$.
  • $N_1$ is open and again we have that $p_2 \in N_1 \cap G_0 \cap G_1$ exists by denseness and this set is open so there exists $0<r_2 < \frac12$ such that $\overline{B(p_2, r_2}) \subseteq (N_1 \cap G_0 \cap G_1)$. Define $N_2 = B(p_2, r_2)$.
  • continue this process recursively.

No distinguishing limit points etc. Just go straight to the goal.

Then the $\overline{N_n}$ $n \ge 1$ form the required nested family that the Cantor intersection theorem can be applied to. The promised $p \in \bigcap_{n \ge 1} \overline{N_n} \subseteq O \cap \bigcap_{n \ge 1}O_n$ witnesses the denseness of $\bigcap_{n \ge 1} G_n$.

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  • $\begingroup$ Quick question, though: how can you be sure that $\overline{B(p_1,r_1)} \subseteq G_1 \cap N_0$, without reference to limit points? $\endgroup$ – Moed Pol Bollo Jul 8 '18 at 1:45
  • $\begingroup$ @MoedPolBollo Because $N_0 \cap G_1$ is open and non-empty, as the intersection of an open and a dense open set. We pick $p_1$ in it. This is an interior point of $N_0 \cap G_1$, so some ball around it sits inside $N_0 \cap G_1$, say $B(p,s_1) \subseteq N_0 \cap G_1$, for some $s_1 >0$. Then take $r_1 = \min(1, \frac{s_1}{2})$ and we have $\overline{B(p_1,r_1)} \subseteq B(p, s_1) \subseteq N_0 \cap G_1$ etc. In any metric space, for an open balls, if $r <s$ we know that $\overline{B(p,r)} \subseteq \{x: d(p,x) \le r\} \subseteq B(p,s)$. It's easy to get smaller closures this way. $\endgroup$ – Henno Brandsma Jul 8 '18 at 5:49
  • $\begingroup$ To the proposer: In any metric space$ (X,d)$ and any $p_1\in X,$ if $0<r_1<r$ then $\overline {N_{r_1}(p_1)}\subset N_r(p_1)$ because we can easily show that $\overline {N_{r_1}(p_1)}\subset \{q:d(q,p_1)\leq r_1\}.$ ... (That is, if $d(p,q)=r_1+s>r_1$ then by the triangle inequality, the open ball $N_{s/2}(q) $ is disjoint from $N_{r_1}(p)$....).... So from the first displayed line of your proof we have $\overline {N_1}\subset N_r(p_1)\subset G_1.$ $\endgroup$ – DanielWainfleet Aug 23 '18 at 4:48
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The proofs of @Bollo and @Brandsma are great and I try to supplement a situation when the metric space is finite.

Supplement proof:

When the metric space $X$ is a finite set, it is always complete. However, there is no such open set $N_r(p_1) \subset G_1$ (in your proof) because every point is an isolated point. Therefore the only dense open subset of $X$ is itself. So the intersection is itself, which is dense.

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