What Makes a Topological Space Second-Countable?

[kent davidge]

I'm taking some study notes on general math, one of the topics being topological spaces.

I've had a look at the definition of what is understandable by countable in Mathematics and I arrived at the following definition for what is a second countable space

- Consider a topological space $(X, \tau)$

- Suppose that $\tau = \cup_{i \ \in \ G \ \subseteq \ \mathbb{N}} \{\tau_i \}$ and that for any $i$, $\{\tau_i \} = \cup_{i \ \in \ F \ \subset \ \mathbb{N}} \{\tau_j \}$. Then $\tau$ has a countable basis and $(X, \tau)$ is second-countable in the topology $\tau$. (Note that $G$ may be the whole of $\mathbb{N}$ whereas $F$ has to be a subset of $\mathbb{N}$.)

Is this definition ok?

 

[member 587159]

I do not comprehend your notations:

First of all, a topological space is denoted with $(X, \mathcal{T})$, so a couple instead of brackets. The brackets make it a set, while it should be an ordered pair.

What does the notation $\mathcal{T} = \cup_{i \in G \subseteq \mathbb{N}}$ even mean?

The definition I would use is:

A topological space $(X, \mathcal{T})$ is second countable if and only if it has a countable basis.

 

[kent davidge]

I do not comprehend your notations:

First of all, a topological space is denoted with $(X, \mathcal{T})$, so a couple instead of brackets. The brackets make it a set, while it should be an ordered pair.

What does the notation $\mathcal{T} = \cup_{i \in G \subseteq \mathbb{N}}$ even mean?

The definition I would use is:

A topological space $(X, \mathcal{T})$ is second countable if and only if it has a countable basis.

Sorry. I've edited my post since then.

 

[member 587159]

Sorry. I've edited my post since then.

Who says that you can index the topology $\mathcal{T}$ with a subset of $\mathbb{N}$? What if the topology contains uncountably many elements? I still think I don't get what you are trying to say.

Just use the definition I gave you. You are making things needlessly complicated!

 

[kent davidge]

Who says that you can index the topology $\mathcal{T}$ with a subset of $\mathbb{N}$? What if the topology contains uncountably many elements?

What I mean is the topology is composed of the open sets $\{\tau_i \}$ their union with the index running over a subset of the naturals or the whole of the naturals is equal to the topology. Is'nt this correct?

Just use the definition I gave you. You are making things needlessly complicated!

I have already seen that definition. But then I would ask what is a countable basis. To put another way, I don't like stopping on that definition because I want a more clear one.

 

[member 587159]

What I mean is the topology is composed of the open sets $\{\tau_i \}$ their union with the index running over a subset of the naturals or the whole of the naturals is equal to the topology. Is'nt this correct?I have already seen that definition. But then I would ask what is a countable basis. To put another way, I don't like stopping on that definition because I want a more clear one.

I still don't quite understand what you mean with your first question. Can you try to write the usual topology on $\mathbb{R}$ as a countable (I.e. indexing set is subset of positive integers) union?

A countable basis is simply a basis that is countable, i.e:

There exists $\mathcal{B}\subseteq \mathcal{T}$ such that

$\mathcal{B}$ is a basis: for every $G \in \mathcal{T}$, there exists $\mathcal{A} \subseteq \mathcal{B}: G = \bigcup \mathcal{A}$

$\mathcal{B}$ is countable: There exists an injection $\mathcal{B} \to \mathbb{N}$

 

[kent davidge]

Can you try to write the usual topology on $\mathbb{R}$ as a countable (I.e. indexing set is subset of positive integers) union?

Well, yes. The open sets are given by $(-i,i)$ where $i \in \mathbb{N}$.

And ya, it seems that we cannot write the usual topology using a subset of $\mathbb{N}$ for the index. Yet, $\mathbb{R}$ is second-countable.

 

[member 587159]

Well, yes. The open sets are given by $(-i,i)$ where $i \in \mathbb{N}$.

No, this is completely wrong. $(-1/2,1)$ is an open set that you didn't include, $(1,2) \cup (3,4)$ is another one.

You gave an open cover of $\mathbb{R}$

 

[kent davidge]

No, this is completely wrong. $(-1/2,1)$ is an open set that you didn't include, $(1,2) \cup (3,4)$ is another one.

You gave an open cover of $\mathbb{R}$

Oh yes, that's right.

 

[kent davidge]

$\mathcal{B}$ is countable: There exists an injection $\mathcal{B} \to \mathbb{N}$

Can you elaborate a bit more on this? The possible injections map the sets in $\mathcal{B}$ into sets in $\mathbb{N}$, correct? How should the mappings be carried out?

 

[WWGD]

Can you elaborate a bit more on this? The possible injections map the sets in $\mathcal{B}$ into sets in $\mathbb{N}$, correct? How should the mappings be carried out?

Consider the endpoints of your intervals, indexing them with a countable set.

 

[member 587159]

Can you elaborate a bit more on this? The possible injections map the sets in $\mathcal{B}$ into sets in $\mathbb{N}$, correct? How should the mappings be carried out?

It doesn't matter how the map looks like. It matters that it exists (or not). And yes, such an injection will map no two different sets in $\mathcal{B}$ to the same natural number.