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.


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?

  • $\begingroup$ 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

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 $\bigcap_{n \ge 1} G_n$.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.