Please comment. (Topology question on separability)

[rumjum]

Homework Statement

If X be a metric space in which every infinite subset has a limit point, then X is separable.

This is a question from Rudin but I am having some difficulty just understanding how to use the hint.

Homework Equations

The hint as in the book is .
Fix delta >0, and then fix E= {xi} such that d(xi,xj) > delta where i not equal to j and xi, xj belong to E. Now show that this process must stop at some time, i.e, the collection of xi's that belong to X such that the distance between them is greater than some delta is finite. Now, choose delta = 1/n (n = 1,2,...) and the center of corresponding neigborhood.

The Attempt at a Solution

Here is my approach & need some help going forward.


Let us choose x1, x2 that belongs to X such that the distance is greater than 1/n. Next let's form a collection An = {xi :d(xi,xj) >= 1/n}. Let's say that there are infinitely many such elements. Now, An is an infinite subset of X and hence should have a limit point.

Let "p" be such a limit point and let us have a neighborhood N1/2n(p) such that N1/2n(p) Intersection "An" has some point xm (that belongs to "An") not equal to p. Now, since "p" is a limit point, and "An" is an infinite subset, the neighborhood of "p" should contain infinitely many points of An. (Or we can chose a radius "r" which is the mininum distance from all points in the neighborhood of "p" and that belong to "An". Then form another neighborhood Nr(p) with the new radius "r" and the intersection of Nr(p) with An shall be null).

Hence, we can choose another such element xm' that belongs to "An". Now, using triangular inequality, we have d(xm,xm') <= d(p,xm) + d(p,xm') <= 1/2n + 1/2n = 1/n. (Contradition) Because , xm and xm' are elemetns of An such that their distance should be greater than 1/n. Hence, there are only finite xi's such that the ditance between them is greater than or equal to 1/n.

Now, each element xj in An is such that we can form a neighborhood Nr(xj) (where r = delta) such that the union of such neighborhoods covers X. Why? If any element "y" of X is not part of An then the d(xj,y) < 1/n, where xj belongs to An. Hence, y should belong to Nr(xj) where r = 1/n.

Hence, there are finitely many such neighborhoods of elements of An that cover X.

Similarly, we can form another such collection An-1, such that the distance between the elements is greater than 1/(n-1). Again this has only finitely many elements. We can repeat this for each delta = 1/n (where n=1,2,...)

Hence, we have A1 subset of A2 (subset of ) A3 ... (subset of ) An ...

Let V = Union of A1, A2, ... An...

[Now this is the part I am getting stuck at].

Consider any Am that belongs to V. Now Am contains finitely many elements that belong to X, such that if p belongs to X, then p belongs to some neighborhood of xm (that belongs to V). Let xm belong to Am, then the neighborhood N1/m(xm) contains p. Hence the neighborhood N1/m(p) shall contain xm. In other words N1/m(p) Intersection V is not null. But V is at least countable as it is union of Ai's , where i = N. And it is dense in X, i.e,any neighborhood of x that belongs to X contains at least one element of V. Hence, X is separable.

(My problem is that I have not used the last line of the hint of the book , "Let delta =1/n, n=1,2... and consider the neighborhoods of the corresponding center".

Please comment.

 

[futurebird]

You need to use $$\delta = \frac{1}{n}$$ to get your countable dense subset. The subset will be the centers of these neighborhoods. Each time you change delta you get a whole new set of neighborhoods.

Let $$\{ N_{\frac{1}{n}}(x_{i}) \}_{n}$$ for (n=1, 2, 3, ...) denote each finite set of neighborhoods for a given vale of n with the radius of the neighborhoods $$\delta = \frac{1}{n}$$. And let,
$$E = \displaystyle\bigcup_{n=1}^{\infty} \{ x_{i}, ... x_{j} \}_{n}$$
be the set of all centers of the neighborhoods for all values of n. E is at most countable because it is the countable union of finite sets. Now you need to show that E is dense in X.