Second countable, countable, proof...

Printable View

February 16th 2011, 02:46 PM
MichaelMath

Second countable, countable, proof...

I have this problem:

S is a set. P is the discrete topology on S. Prove that (S,P) is second countable iff S is a countable set.

Can somebody expand on this? (I assume it's basically what I'm trying to prove):

"Every discrete space is first-countable; it is moreover second-countable if and only if it is countable."

Is this way of thinking correct (I'm an undergrad):

A topology is second countable if it has a second countable base.

A base is second countable if it contains a countable number of open sets.

The discrete topology on S is the power set of S, where, by definition, every subset of S is open.

Am I on the right track? Do I just say something along the lines of S being a set of open balls and show that it's countable?

(Wondering)
February 16th 2011, 03:47 PM
Tinyboss

Second countable means the topology on the space has a countable basis (you don't say "second countable basis"). In the discrete topology, the set of singleton sets is the smallest basis.
February 16th 2011, 05:32 PM
MichaelMath

would that be made up of the centres of all the balls in S?
February 16th 2011, 05:36 PM
Tinyboss

In the discrete topology, every subset is open, and so the smallest basis is the set $\big\{\{s\}\mid\:s\in S\big\}$ . Every subset can be written as a union of points (actually singleton sets).
February 17th 2011, 05:59 AM
MichaelMath

Sorry, I kind of get what you are saying, but I'm not quite sure how to go about proving this:

Quote:

"Every discrete space is first-countable; it is moreover second-countable if and only if it is countable."
February 17th 2011, 06:52 AM
Plato

Quote:

Originally Posted by MichaelMath

Every discrete space is first-countable; it is moreover second-countable if and only if it is countable."

Suppose that $(\mathcal{S},\mathcal{T})$ is discrete topological space. i.e. $\mathcal{T}=P(\mathcal{S})$ .
Then for each $s\in\mathcal{S}$ the set $\{s\}$ is an open set.
Therefore, at each point there is a local basis at the point, first countable.

Now if the discrete space were second countable then there could only be countably many singleton sets. So the space itself is countable.
February 17th 2011, 09:40 AM
MichaelMath

deleted
February 17th 2011, 09:42 AM
MichaelMath

Is this excessive?

$S$ is a discrete topological space if every subset of $S$ is open. In particular, for each $s \in S$ we have $\{ s \}$ is an open set.

$S$ is first-countable, if for any $s \in S$ , there is a countable sequence $\{ U_n(s), n>0 \}$ of open sets containing $s$ , such that for any neighborhood $V$ of $s$ there exists a member $U_i(s)$ of this sequence, such that $U_i(s)$ is contained in $V$ .
Given a neighborhood $V$ of $s$ , we have that $s \in V$ . But $\{ s \}$ is an open set. So, we can choose the sequence $\{ U_n(s) \}$ to consist of just one set $\{ s \}$ . For every neighborhood $V$ of $s$ we then have an open set $\{ s \}$ contained in this neighborhood.

$S$ is second-countable, if its topology has a countable base. More precisely, there's a countable collection $\{ U_n, n>0 \}$ such that any open set $V \in S$ can be written as a union of some (finite of infinite) number of elements of $\{ U_n \}$ .

Suppose $S$ is discrete and countable. Then, since every one-point set is open, any $V \subset S$ can be written as a union of $\{ s_i \}$ , where $\{ s_i \in V \}$ .

Suppose, $S$ is second-countable and discrete. Let $\{ U_n, n>0 \}$ be a countable collection of open sets with the property that every open subset $V$ on $S$ can be written as a union of some members of $\{ U_n, n>0 \}$ .

Assume that $S$ is uncountable and get a contradiction:

Let $s \in S$ . Since $\{ s \}$ is an open set, $\{ s \}$ can be written as a union of some elements of $\{ U_n \}$ . But the only way to write a one-point set as a union of two sets is to say $\{ s \} = \{ s \} \cup \emptyset$ Thus, the set $\{ s \}$ belongs to $\{ U_n \}$ . Since this should be true for all $s \in S$ , and $S$ is uncountable by our assumption, we conclude that $\{ U_n \}$ contains uncountably many open sets $\{ s \}$ . Contradiction. We conclude that $S$ must be countable.
February 17th 2011, 10:26 AM
Plato

Quote:

Originally Posted by MichaelMath

Is this excessive?

Yes it is excessive.
Because a collection of one is a countable collection.
February 17th 2011, 11:02 AM
MichaelMath

Thought so, but it was the only other related explanation I could find. Do you reckon your previous post is a sufficient proof?
February 17th 2011, 11:10 AM
Plato

Quote:

Originally Posted by MichaelMath

Do you reckon your previous post is a sufficient proof?

I have no way of knowing that.
I do come from a tradition in proofs were written in whole sentences and paragraph form. We were expected to use as little notational short cuts as possible.
February 17th 2011, 11:21 AM
MichaelMath

Quote:

Originally Posted by Plato

Because a collection of one is a countable collection.

Can you expand on this?
February 17th 2011, 12:06 PM
Plato

Quote:

Originally Posted by MichaelMath

Can you expand on this?

What does first countable mean?
There is a countable local basis at each point.
For each $p\in \mathcal{S}$ the singleton set $\{p\}$ is a basis element.
That is a countable local basis. For any open set $\mathcal{O}$ if $p\in \mathcal{O}$ then $p\in\{p\}\subseteq\mathcal{O}$ .
February 18th 2011, 07:08 AM
MichaelMath

Quote:

Originally Posted by Plato

Suppose that $(\mathcal{S},\mathcal{T})$ is discrete topological space. i.e. $\mathcal{T}=P(\mathcal{S})$ .
Then for each $s\in\mathcal{S}$ the set $\{s\}$ is an open set.
Therefore, at each point there is a local basis at the point, first countable.

Now if the discrete space were second countable then there could only be countably many singleton sets. So the space itself is countable.

Could you tell me how this follows from the previous statements? I don't see the logical link.