Vtyjkyuk

Let $G_n$ be a sequence of dense open subsets of a complete metric space $X$. Show that $ bigcap_n geq 1G_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

$overlineN_1 subset G_1.$

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 $r_2 < minbig(r_1, frac12big)$ such that
$$overlineN_2 subset overlineN_1$$
and, since the $G_n$ are all open, we also choose $r_2$ small enough that
$$overlineN_2 subset G_2.$$

Continuing this process, we can construct the sequence of closed subsets $BigoverlineN_nBig$. Further, by the way we have chosen the $r_n$ (i.e., $0<r_n<frac1n$), it follows $r_n to 0$, so we have actually constructed a nested sequence of closed bounded sets $BigoverlineN_nBig$ such that $diam overlineN_n to 0.$ Since $X$ is a complete metric space, the intersection of this sequence is nonempty and since each $$overlineN_n subset G_n$$ 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?

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 $overlineB(p_1, r_1) subseteq G_1 cap N_0$. 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 $overlineB(p_2, r_2) subseteq (N_1 cap G_0 cap G_1)$. 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 $overlineN_n$ $n ge 1$ form the required nested family that the Cantor intersection theorem can be applied to. The promised $p in bigcap_n ge 1 overlineN_n subseteq O cap bigcap_n ge 1O_n$ witnesses the denseness of $bigcap_n ge 1 G_n$.

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.