Why defining seperable topological space we use "COUNTABLE" dense set?

Question

My question is the caption itself. Thanks for helping.

Answer 1

It's just a clever definition where the key appears to be taking a small number of points as being dense. For one thing a larger number of points has an easier time being dense... I guess one might try replacing countable with finite, and look at a smaller class of spaces. But to make things more interesting, countable was used in the definition. Of course countable is the smallest level of infinity. A countable set can be put in a list. In the case of a separable topological space, I guess you could almost think of the points being partitioned, if you will, or separated by the subset.

If you went in the other direction and said, instead of countable, $\aleph_3 $, or $\aleph_{1000}$, or higher cardinality yet, you would have something a little more obscure and hard to get a feel for... though still interesting perhaps. ..

Of course, it's a definition... When you define something, you can be creative and define whatever you want, quite apart from whether such things even exist, or proving theorems about them...

Answer 2

It is convenient to be able to name each point in a space. If the space is countable, then there is no difficulty in doing so. If the space is uncountable, then attaching a name to each point is impossible. However, if a countable dense set is present, then you can name each point in the space by a sequence from that countable set. This gives you a rather simple mechanism to be able to 'reach' every point in the space.

In more technical terms, having a countable dense set allows for recursive constructions and inductive proofs in the space without the need for any transfinite arguments past $\omega$.

Answer 3

First reason: it is a useful definition. If a space $X$ is separable and we want to prove some properties of it, we can first work on the countable subset $C$ which is much easier to handle. Afterwards we can take the closure of $C$, i.e. limits.

Second reason, omitting the condition "countable" makes the definition useless, because every space is dense in itself.