There are people posting on Veeky Forums RIGHT NOW that can't prove that every compact metric space has a countable...

countable dense subset?
More like, noooooooo noooo nooooo

>there are people posting on Veeky Forums RIGHT NOW dumping their questions outside of the SQT

CA

Assume there exists an unseparable compact metric space X. Now due to X being unseparable, we can construct an uncountable collection U of nonempty separate open balls with radius r in X. Now there exists a sequence (x_n) s.t. each element is placed in a different open ball, where each open ball belongs to the collection U. Clearly the sequence (x_n) doesn't have converging sub-sequences, thus X is not compact.

can you do one that isn't a proof by contradiction?

for example, if you take a closed interval, you can moved arbitrarily close to any point by continually dividing halves like an infinitely deep binary tree. i don't know how to express this formally though or in a more general way.

>just take the centers of your balls, that's a countable dense set

>just take the centers of your balls
but that is what I'm doing, if I don't take a limit first it won't be dense and for the limit to be well defined I needed that condition. Also, how can you get an uncountable limit this way, you got an example?

what does it mean to take the limit of a sequence of sets?

>Let B_n be a finite subset of A_n which is an open covering of X (X compact so this exists)

do you mean "Let B_n be a finite subset of A_n *that* is an open covering of X"?

gotta watch out for errors like that.

>if I don't take a limit first it won't be dense
it will
>for the limit to be well defined I needed that condition
which limit? you have an infinite number of sequences branching around

here's a good example to show you how weird your construction is. take the compact metric space to be [0,1]. your construction gives, say, the rationals {1/2}, {1/3,2/3}, {1/4,2/4,3/4} ...
you can see this (countable) set is dense in [0,1] and the limits in this case are the whole space

I think I meant "which" but with a comma after A_n, but I am not a native speaker.

In this case we take a countable union of finite sets, which is most definitely countable, even countable unions of countable sets are, so I don't see the issue.

using "which" reads like an assertion that any finite subset B_n is an open covering of X, whereas "that" constrains B_n to be an open covering. just a small grammatical issue.

>which limit?
the limit I constructed
>
here's a good example to show you how weird your construction is. take the compact metric space to be [0,1]. your construction gives, say, the rationals {1/2}, {1/3,2/3}, {1/4,2/4,3/4} ...
you can see this (countable) set is dense in [0,1] and the limits in this case are the whole space
This is dense but my construction wouldn't allow this example and the limit also wont give us the whole space in my construction, we are taking the union of sets here so x in B means there exists an n such that x is in B_n, which we wouldn't have for, say, the irrationals if we decide to always pick rational numbers when choosing our finite coverings.

saying "limit" instead of just "union" is sorta weird but ok

It is a limit of the sequence B_n, but also a union

>This is dense but my construction wouldn't allow this example and the limit also wont give us the whole space in my construction
>This can't be my construction because it's wrong and my construction is right
Well.

the limit of those sequences is NOT the same as the union of the centers. the union of the centers works and is correct, taking all limits gives the whole space

You are forgetting that I assumed B_n-1 subset B_n.
Otherwise you would be correct.

it can't be my construction because you don't have B_n-1 subset B_n

>Let A_n be the set of open balls around every point x in X of radius 1/n, A_n is obviously an open covering of X. Let B_n be a finite subset of A_n which is an open covering of X (X compact so this exists) and we assume wlog B_(n-1) is contained in B_n (all the B_n are finite anyways so we can do this).

>take a finite 1/n-net for each natural n, and the union of them is a countable dense set

is there a neater way to do this in general? since it's a metric space, can you say something like "here's a set of n points that are 'evenly distributed' or not too close together"?

What do you mean by "in general"? I think I read that this also holds true for pre-compact metric spaces, don't ask me for a proof though.

for an interval, the example in
uses an infinite binary tree where each node represents a fraction. each depth level of the tree has a finite number of nodes, and taking the finite union gets you a dense countable subset.

somehow this seems nicer to me than taking the countable union over n of the centers of the balls in the B_n's as defined in
the construction just has to make sure they're "spaced out" properly, so that the countable union forms a dense set, right?

>finite union

*countable union

Proof by basic introductory topology.

Let [math]A_n = \def{\{}B_x | B_x = B(x,\frac{1}{n}), x \in K \def{\}}[/math]
Because [math]K[/math] is compact, there exists a finite [math]A_{n'} \subset A_n[/math] that covers [math]K[/math]
Now, let [math]B_n[/math] be a collection of the central points of every ball in [math]A_{n'}[/math]
It should be clear that [math]\cup_n B_n[/math] is now both countable and dense in [math]K[/math]

[math]A_q[/math]

[math]A_Q [/math]

[math]A_Q_R_t_u_V_W_X_Y_z [/math]

...

correct

bless you