proofread proof that if every sequence in metric space K converges to some point of K, then K compac

[tidal bay]

Hi I wrote this proof to try to convince my professor that im ready for real analysis and he should let me take the class. is the proof overall good enough and clear enoguh? Am i insane and is it all wrong? can someone please with a cherry on top help

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[tidal bay]

Ill paste it here too

or try to

[subtle spire]

why is infinity of the subset required?

being nonempty should suffice

you make it look like you need the elements in a sequence to be distinct

why can such x_{n+1} be picked?

you're saying "pick elements such that they are far away from all the previously chosen elements"

why is this possible?

???

describe more precisely what E is

why is E nonempty for starters?

These arguments are very confusing to read. I think your setup goes something like this:

1. K is sequentially compact, therefore K is separable
2. Let E be countable everywhere dense subset of K
3. Let S be the collection of open balls with rational radii and centers in E. S is a countable set.

Let V_a be an arbitrary open cover of K. Let T be the subcollection of S such that every element of T is a subset of at least one of the V_a. T is countable.

@tidal bay

the progression works, though, we do show that V_a has a countable subcover, which we then further refine to a finite subcover, proving compactness

[wispy ember]

actually why does E need to be finite?

what about (x_n) : 1,2,1,2,...

(noticed aL already said this mb)

[tidal bay]

thats what I had before im not sure why I changed i

[tidal bay]

OK gimme a minute i'll rewrite this section more clearly

The idea is to use the fact that every infinite subset has a limit point to prove that X can be covered by finitely many neighborhoods of radius delta, by proving that at some point you run out of such x

Also E should contain all x_n

[subtle spire]

all x such that..?

[tidal bay]

x_n they are theh ones we just picked

I missed the _n

when I compiled the latex

[tidal bay]

yeah I know, sorry, this is mostly inspired by a bunch of rudin exercises I worked through previously so I proved it assuming they were true, then in an actual proof id have to include all the logic

[tidal bay]

because you pick any random x_1

if the entirety of E is covered by one neighborhood of radiius delta, thahts it

otherwise its possible to select more x_n

[subtle spire]

ok, so the argument is if E is infinite, then it must have a limit point, hence by definition of limit only finitely many elements can be far away from it

[tidal bay]

yeah

[subtle spire]

you are saying tho that E is the set of all such x_n

[tidal bay]

yes

[subtle spire]

a limit point of E need not be an element of E

[tidal bay]

I just assumed that the limit point was any point

[subtle spire]

nono, that wont fly

[tidal bay]

assume it is not an element of E?

or is

wait

first we finish making E

[subtle spire]

this E seems superflous, lemme think

[tidal bay]

then for contradiction we assume that E is infinite

means it must have a limit point

this limit point creates a contradiction

[subtle spire]

your argument goes like this:

cover K with delta balls centered at each x in K

then K must be covered by finitely many such balls

[tidal bay]

yes

because every infinite subset of K has a limit point

and the set of all centers of such balls cannot have a limit point

otherwise we would get the aforementioned contradiction

[subtle spire]

ok, so if only infinite covers were possible, then we pick one such cover and call the set of all the center points E

then it must have limit point in K, say y, so there's a sequence of elements in E converging to y in K

but E consists of center points of the fixed infinite cover, since y is a limit, only finitely many elements in the sequence are far away from y, so we can reduce to a finite cover, a contradiction

ok i think this works

every infinite subset has a limit point in K because we can pick an arbitrary sequence and by assumption it must have a convergent subsequence with a limit in K

[tidal bay]

this is the exercise in rudin

[tidal bay]

yeah that does work

I wrote this before sequenecs

that part

[subtle spire]

ok so we got most of that done

there's the last sentence to get a countable dense subset

fix delta = 1/m, then from the previous argument we get a finite cover of radius delta, call the center points like

$$x_1^m,\ldots, x_{N(m)}^m$$

[patent lake]

aL

[subtle spire]

and you can define

$$Q := \bigcup {m\in\mathbb N} {x_1^m,\ldots, x{N(m)}^m}$$

[patent lake]

aL

[subtle spire]

in summary: you pick a finite cover for every m in N and then you take the union of their center points

this is clearly dense and countable

[tidal bay]

yeah you take all centerpoints for all 1/n, and for a neighborhood of x with radius r just take 1/n < r and the centerpoint of the neighborhood with radius 1/n containing x

so its dense

and countable

[tidal bay]

yeah thats a better way to define E probaly

[subtle spire]

1. is covered now.

[subtle spire]

all that remains is to prove these two

[subtle spire]

emphasise that E_delta is a finite set

the ellipsis is confusing the reader

[subtle spire]

better, but your n depends on delta

n = n(delta)

[subtle spire]

yes, you can also explain the notation a bit more for the reader

[tidal bay]

perhaps "let their centers be E"

[subtle spire]

it's very good right now

but emphasise that the number of elements in E_delta, also depends on delta, hence the notation n(delta)

[woven imp]

Congratulations @subtle spire, you have been awarded the <@&1257594408103317575> for being the most active user today.

[subtle spire]

fml..

[upbeat otter]

@tidal bay

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