Verified answer

Fix a metrizable space X and a metric d on X that gives the topology. Fix also a closed set S in X. For x in X, define

d(x,S) = inf {d(x,s): s in S}

and define for n = 1, 2, 3, ... the subsets

S_n = {x in X: d(x,S) < 1/n}

of X. I claim (1) for all n the set S_n is open, and (2) S is the intersection of all of the sets S_n. These two facts together show that S is a G-delta set.

Fix n and let's show that S_n is open. Fix y in S_n. By definition of S_n we have d(y,S) < 1/n, so there is s in S with d(y,s) < 1/n. The number e = 1/n - d(y,s) is positive. I claim the open ball in X with center y and radius e is entirely contained in S_n. But this is clear: if x in X satisfies d(x,y) < e then

d(x,s) <= d(x,y) + d(y,s) < e + d(y,s) = 1/n - d(y,s) + d(y,s) = 1/n

and hence d(x,S) < 1/n, and hence x in S_n, as desired. This shows that S_n is open.

It is clear that if t is in S then d(t,S) = 0 (because the set {d(t,s): s in S} includes the number d(t,t) = 0) and hence that t is in each of the sets S_n. This shows that S is contained in the intersection of the sets S_n. Conversely suppose that x in X is in all of the sets S_n. Construct a sequence y_n in S as follows: fixing n, from the fact that x is in S_n we know d(x,S) < 1/n, and hence that there is some element y of S with d(x,y) < 1/n; choose one and call it y_n. The sequence thus defined satisfies d(x,y_n) < 1/n for all n. It follows that the sequence (y_n) converges to x (since if e > 0 is given, choosing N so that 1/N < e, one has that d(x,y_n) < 1/n < 1/N < e for all n >= N) and hence that x is in the closure of the set S, and since S is closed, that x is in S. This concludes the proof.

The converse is not true although I had to go to the web for examples. Wikipedia tells me that spaces where every closed set is G-delta are called "G-delta spaces". It lists various properties of these spaces and provides an example of a space, the Sorgenfrey line, which is a G-delta space but not metrizable.

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