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.

Expert's answer

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>0.

On the other hand, the sequence

(x_n) converges to x, so there exists N>0 such that x_n belogs to U for all

n>N.

But then x_n with n>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>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.

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>0.

On the other hand, the sequence

(x_n) converges to x, so there exists N>0 such that x_n belogs to U for all

n>N.

But then x_n with n>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>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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