# Answer to Question #7517 in Real Analysis for Jim

Question #7517
Prove: A subset S of the real numbers is closed if and only if lim (as n goes to infinity) X_n is an element of S whenever (X_n) is a convergent sequence whose terms are all in S.
1
2012-03-20T10:52:33-0400
By definition a subset S of R is closed if its complement R\S is open.

1)
Suppose S is closed, so R\S is open.
Let (x_n) be any sequence of elements of
S converging to some x.
We should show that x belogs to S.

Suppose x
belogs to R\S.
Since R\S is open, there exist an open interval U=(x-a, x+a)
that does not intersect S for some a&gt;0.
On the other hand, the sequence
(x_n) converges to x, so there exists N&gt;0 such that x_n belogs to U for all
n&gt;N.
But then x_n with n&gt;N belogs to R\S, which contradicts to the
assumptino that x_N belongs to S.

Hence x belongs to S.

2)
Conversely, suppose S is not closed, and so R\S is not open.
We will find a
sequence of points from S converging to some x belonging to R\S.

Notice
that the assumptino that R\S is not open means that there exists x from R\S such
that for any a&gt;0 the
interval (x-a, x+a) contains a point y_a from
S.
Consider the sequence x_n = y_{1/n}.
Since 1/n converges to 0, it
follows that x_n converges to x.
Thus we obtained as sequence of points from
S converging to some x belonging to R\S.

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