# Countable base and subcover

1. Jul 21, 2010

### wayneckm

Hello all,

I am a newbie to topology. Hope someone can help sharpen my understanding.

I read a theorem that "if $$T$$ is a topological space with countable base, then every open cover has a finite or countable subcover"

Apparently this is quite intuitive because as in the definition of countable base, every open set should be generated by it, and $$T$$ is by definition open, so that means it should be the union of the countable base. Obviously this is the "smallest size" open cover which is of course countable, so any open cover cannot get smaller "size" than this?

Also, somehow I am quite confused with countable subcover, because if an open cover is "as countable as" the base, this means every set is just that in the base, hence, this is the just the open cover formed by the base, so is the above theorem valid, i.e. open cover and subcover are the same is allowed?

Thanks.

2. Jul 21, 2010

### quasar987

Well, yes, that is the correct idea, but you should be able to show how to construct a countable subcover from a given uncountable cover {O_i}.

This I don't follow. Maybe you can rewrite that question in other (more precise) terms?

3. Jul 22, 2010

### wayneckm

Actually I am confused with the following cases:

i) An uncountable open cover ----> countable subcover (Meaningful)

ii) Countable open cover ----> countable subcover (Meaningful, because the former one can be of cardinality larger than the later one, so subcover makes sense to me)

iii) Countable open cover ---->?? countable subcover (if they have the same cardinality, or in other words there exists homeomorphism, I do not know if I use correctly the word "homeomorphism" here, so in this case apparently we cannot further extract a subcover)

So in general, we should be interested in reducing the cardinality of an open cover while maintaining the "covering ability" when we talk about subcover?

Thanks again!

4. Jul 22, 2010

### quasar987

What does "---->" mean here? And in what sense are you using the word "Meaningful"? Also, ii) and iii) look the same to me, so why is ii) meaningful and iii) is not?

Huh???

5. Jul 22, 2010

### wayneckm

Sorry for my unclear presentation.

"Meaningful" to me means I can get a "strictly smaller" collection of sets from its original collection of sets.

So in (ii), I assume the open cover contain "more" sets while still being countable, and the subcover is formed by throwing away at least one set from the open cover. So the "size" is strictly smaller.

In (iii), I assume the open cover is already the "smallest" one, so in this case, I don't know whether there exists a subcover.

Or back to the first post, if the open cover is the union of countable base, can we still make the same conclusion?

6. Jul 22, 2010

### Landau

I am sorry, but you are still not making any sense. If S is a countable (infinite) set and s is an element of S, then both S and S-{s} are countable; you cannot say that S-{s} has "strictly samaller size" unless you have defined some weird notion of size which I don't know. Also, you haven't explained what you mean by "--->". Please try to use precise, correct mathematical language.

7. Jul 22, 2010

### wayneckm

So sorry again. Maybe I should write in this way:

if $$T$$ is a topological space with countable base, then every open cover has a finite or countable subcover.

If I take the open cover to be union of elements in countable base, then it covers $$T$$. So is there any finite/countable subcover in this case?

8. Jul 22, 2010

### quasar987

Ok, and what is the definition of a subcover? If you read the definition carefully, you will notice that every open cover is a subcover of itself.

So if B is your countable basis, then, yes it admits a countable subcover, namely, B itself.

This is the trivial case.

The real problem is in showing that if you have an uncountable open cover, then you can find a countable subcover.

9. Jul 22, 2010

### wayneckm

Alright, that's what I need. Thanks so much!!!