/sci/ - Science & Math

bump.

It only holds for sequences of non-empty intervals.

And please don't use \epsilon when you mean \in

>>6212172
So what I said was correct.
<span class="math"> x\epsilon X[/spoiler]
<span class="math">x\in X[/spoiler]

let's see the difference... I really don't know. Also, I'm new to TeX stuff

>>6212181
Wow. You're right!>>6212172

>>6212181
>So what I said was correct.
No.

>>6212172
empty interval... give me a break.

>>6212154
>For <span class="math">\cap I_{k}[\math] to be empty, there must exist a <span class="math">I_{k}= \emptyset[\math].
You're either assuming what needs to be proved, or else assuming that any decreasing chain
of sets whose intersection is empty must have an empty element, which is false.[/spoiler][/spoiler]

But look at this. I_n = (0, 1/n). Take the intersection of them all, and you have nothing, yet each is nonempty and they are nested. So there must be a flaw in your proof. Note that my intervals are open.

>>6212198
[1,-1] is an empty interval.

>>6212200
>>6212199
WTF is this? A raid from retardchan? Do you even know what a closed interval is?

>>6212192
What did I say that was incorrect?

I think the proof is flawed. Suppose you had open intervals instead: (1-1/n,1) these are nested, and the intersection is empty, but no interval is empty.

>>6212200
Note that the problem asks about closed intervals.

>>6212208
yes, but the deductive step that is key to OPs proof is not clear. The same logic he uses would apply to open intervals and the result is false in this case.

>>6212209
What does that have to do with closed intervals? In agree that with open intervals you'd be right, but not closed.

>>6212206
>>6212208
My point was that his proof does not use the closedness assumption. So it must be wrong. I am well aware that decreasing compact sets intersect nontrivially.

>>6212206
>>6212208
OP assumed that any decreasing chain of sets whose intersection is empty must have
an empty element, which is obviously false as >>6212200 showed.

>>6212209
>>6212213
>>6212216
Are you fucking braindead? OP didn't prove anything. He provided a counter-example in the case that the statement doesn't require the intervals to be non-empty. If we were on stackexchange, you would have 20 downvotes by now.

>>6212218
Are you reading impaired? What kind of brain damage did you suffer? OP showed that an empty set in the chain would destroy the statement.

>>6212220
Good grief. I mean... Good grief.

Do you want a proof of this OP or do you just want us to point out what's wrong in yours?

>>6212229
There is nothing wrong in his. The only thing that is wrong is your reading comprehension (or more precisely the lack thereof).

Lol. The confusion here arises from the fact that OP also forgot to state that the intervals should be nonempty. I don't remember exactly how Munkres defines intervals. Perhaps they are nonempty by definition. In any case, this should be assumed in the exercise. The point of it is to expose an interesting fact about nested closed sets.

>>6212154
>For <math> to be empty, there must exist a <math>
Why is that, OP? Someone else already demonstrated that this implication does not hold for general nested sets. How does this follow from the properties of nested closed intervals?

>>6212227
I share your pain.

>>6212238
Proof is wrong even if intervals are defined to be non-empty.

>>6212251
Again: It was never meant to be a proof, you illiterate cretin.

>>6212254
Then what was it meant to be? It contains the statement "For <math> to be empty, there must exist a <math>", which is demonstrably false. What was that all about?

>>6212254
Psst! Everyone's ignoring you!

>>6212254
OP is asking if his proof by contradiction is valid. It isn't.

[n, inf) are closed intervals. They are nested and nonempty. Intersection of them all is empty.

OP gets your fucking shit together. By interval you mean a closed and bounded interval. Then the exercise makes sense and has a simple proof using completeness.

>>6212261
It is very valid.

>>6212264
It's a pretty standard convention to require intervals to be bounded.

>>6212289
No! It isn't!
He assumes that every decreasing chain of sets with empty intersection
must have an empty element, which is false.

>>6212305
>He assumes that every decreasing chain of sets with empty intersection must have an empty element, which is false.

Show me a decreasing nested chain of non-empty compact intervals in R with empty intersection.

>>6212305
Why is it false?

>>6212312
>non-empty compact intervals
I didn't say that.
Obviously he can't assume this either because it's precisely what he's trying to prove.
You do realize that circular proofs are invalid, right?

>>6212315
See >>6212200

>>6212316
But those intervals are open and not closed.

>>6212316
He didn't attempt a proof. He only demonstrated why a "non-empty" requirement is needed. What is your IQ? Why can't you into reading comprehension?

>>6212316
>>6212200
>OP talks about closed intervals
>anon: hurrrrrrrrrrrrr look at muh open intervals

A new all-time low for /sci/.

>>6212319
No shit. Your point?

>>6212322
> He only demonstrated why a "non-empty" requirement is needed
maybe you should tell that to OP, because there's nothing about his post that indicates that

>>6212328
>Your point?

My point is that you're too retarded to understand the problem. Perhaps you should take a math class before further embarrassing yourself.

>>6212328
What elementary school did you attend where the teachers failed to teach you basic literacy? OP clearly states what he's doing. He proves that the statement is wrong if you don't require the intervals to be non-empty, you fucking retard.

What's with all the trolling ITT? Nobody can seriously be that retarded not to understand what OP wrote.

>>6212326
>>6212333
So, basically, what you insipid fuckholes are saying is that it's logically acceptable to use that fact that a decreasing chain of non-empty compact intervals has a non-empty intersection in order to prove that, you know, a decreasing chain of non-empty compact intervals has a non-empty intersection....

>>6212336
See >>6212260

>>6212348
Where did he ever attempt to prove the statement, you illiterate piece of shit? All he did was demonstrating why it is wrong without the "non-empty" requirement. You are fucking retarded. Please stop posting. Your toxic lack of intellect is detrimental to this board.

>OP: Without demanding non-emptiness the theorem becomes wrong and here is why ...
>anon: hurrrrrrrrrrrr that's not how you prove the theorem durrrrrrrrrr

the retardation ITT has reached critical mass

>>6212348
Why would he prove a wrong statement when he can just provide a counter-example like he fucking did?

>>6212348
>is called out for his stupidity
>"lol I'm ignoring you"

You are wrong on a math board.

OP: Please check my proof
Rational People: OP's proof has flawed logic.
Troll: OP didn't offer a proof fucking morons!

I love this board.

>>6212154
With regards to the picture, doesn't a toad have the same topology as a human, and a wizard's hat, and a thimble? Since there are no "holes" in the surface of any of them?
(I'm practically below babby tier when it comes to topology)

>>6212528
frogs and humans have the same topology as a sphere.
Thimbles and Wizards hats have the same topology as a disk.

>>6212496
0/10

>>6212496
>OP: Please check my proof

Except he never said that. Reading comprehension must be really hard.

>>6212573
I'm a torus at least; dunno about you

>>6212528
>buttplug

>>6212573
> sphere
Most humans have at least 1 hole m8

>>6212154
I went all the way to university last week during the holidays to borrow Munkres only to find it was already out. Fucking bet it was you OP

>>6212973
Get a fucking pdf. Or buy it. 100% worth owning a copy of Munkres, and you can get it quite cheaply.

>>6212972
I'm no biologist, but I count three true "holes" preventing simple connectedness - digestive tract all the way through, and the nostrils connecting to the mouth/sinuses. Any biologists want to shed more light on this?

>>6212154
First, every $ I_k $ must be bounded and non-empty.
Every interval will be like this <span class="math"> [a_n, b_n] [\math], since <span class="math"> a_n [\math] and b_n [\math] are monotone and bounded sequences, both converge. Take the sequence x_n = \frac{a_n+b_n}{2} [\math] the limit will be an element from \bigcap_{n \in N} I_n [\math] because it is in every I_k [\math].

Without sequences, for any compact non-empty subsets.
Take G_k = I_k^c [\math] , if \bigcap_{k \in N} I_k = \emptyset [\math], \bigcup_{k \in N} G_k = R [\math] and it's an open cover of every I_k [\math]. Since I_a [\math], a \in N [\math] is compact, there is a finite subcover such that G_{a_1} \cup G_{a_2} \dots \cup G_{a_n} = I_a [\math]. But this means that I_a \cap I_{a_1} \cap I_{a_2} \dots \cap I_{a_n} = \emptyset [\math], which is a contradiction.[/spoiler][/spoiler]

>>6212154
Every <span class="math"> I_k [/spoiler] must be bounded and non-empty.
Every interval will be like this <span class="math"> [a_n, b_n] [/spoiler], since <span class="math"> a_n [/spoiler] and b_n are monotone and bounded sequences, both converge. Take the sequence <span class="math"> x_n = \frac{a_n+b_n}{2} [/spoiler] the limit will be an element from <span class="math"> \bigcap_{n \in N} I_n [/spoiler] because it is in every I_k .

Without sequences, for any compact non-empty subsets.
Take <span class="math"> G_k = I_k^c [/spoiler] , if <span class="math"> \bigcap_{k \in N} I_k = \emptyset [/spoiler], <span class="math"> \bigcup_{k \in N} G_k = R [/spoiler] and it's an open cover of every <span class="math"> I_k [/spoiler]. Since <span class="math"> I_a [/spoiler], <span class="math"> a \in N [/spoiler] is compact, there is a finite subcover <span class="math"> G_{a_1} \cup G_{a_2} \dots \cup G_{a_n} = I_a [/spoiler]. But this means that <span class="math"> I_a \cap I_{a_1} \cap I_{a_2} \dots \cap I_{a_n} = \emptyset [/spoiler], which is a contradiction.

>>6213442
>the limit will be an element from [...] because it is in every I_k .
>the limit
There is no reason such a limit exists.

I haven't yet looked at your alternative proof, but I suggest this: Since the assertion trivially fails if any interval is empty, let's assume every interval is non-empty. This means there exists a sequence <span class="math">a_n, n \in \mathbb N[/spoiler] with <span class="math">a_n \in I_n[/spoiler]. Then by construction <span class="math">a_n \subset I_1[/spoiler], and since the interval is closed (and, I presume, bounded) it is also compact, and (by the Bolzano Weierstrass theorem) there exists a sub-sequence <span class="math">a_{n_i}, i \in \mathbb N[/spoiler] of <span class="math">(a_n)[/spoiler] which converges to some limit in <span class="math">I_1[/spoiler], <span class="math">lim_{i \rightarrow \infty} = a \in I_1[/spoiler]. Further, for every <span class="math">n \in \mathbb N[/spoiler], the tail sequence <span class="math">a_\nu, n \le \nu[/spoiler] starting at <span class="math">n[/spoiler] is contained in <span class="math">I_n[/spoiler]. Hence there also exists a tail of the sub-sequence <span class="math">(a_{n_i})[/spoiler], which is also contained in <span class="math">I_n[/spoiler] (for instance, starting the tail at the n-th element, <span class="math">a_{n_\iota}, n \leq \iota[/spoiler] will do the trick because <span class="math">n_i \geq i[/spoiler] always holds by the nature of sub-sequence indices). Then the tail of the sub-sequence is contained in <span class="math">I_n[/spoiler], and by compactness of the latter also assumes its limit <span class="math">a[/spoiler] in that interval, so <span class="math">a \in I_n[/spoiler] holds for all <span class="math">n \in \mathbb N[/spoiler], and hence <span class="math">a \in \bigcap_{n \in \mathbb N}[/spoiler], so the intersection is not empty.

>>6213380
>there are undergrads on /sci/ that do not own the 10 dollars indian edition of Munkres topology.

http://www.abebooks.co.uk/servlet/BookDetailsPL?bi=4558009077

>>6213507

You have to use paizano's constans.

>>6213514
Why would I want that? I don't need a book full of trivialities I already know.

>>6213442
>there is a finite subcover <span class="math">G_{a_1} \cup G_{a_2} \dots \cup G_{a_n} = I_a[/spoiler]. But this means that <span class="math">I_a \cap I_{a_1} \cap I_{a_2} \dots \cap I_{a_n} = \emptyset[/spoiler], ...
No. In the first equation the set-relation is wrong, fix it to <span class="math">G_{a_1} \cup G_{a_2} \dots \cup G_{a_n} \superset I_a[/spoiler]. Next, by means of taking complements (which I think is what you intended) this is equivalent to <span class="math">I_a \cap I_{a_1} \cap I_{a_2} \dots \cap I_{a_n} \subset G_a[/spoiler].

>which is a contradiction.
Besides from presumably being miscalculated this does not contradict your assumption which reads <span class="math">\bigcap_{k \in N} I_k = \emptyset[/spoiler] and actually is entailed by your last formula of <span class="math">I_a \cap I_{a_1} \cap I_{a_2} \dots \cap I_{a_n} = \emptyset[/spoiler].

>>6213643
(and once more -- fixing \supserset)
>>6213442
>there is a finite subcover <span class="math">G_{a_1} \cup G_{a_2} \dots \cup G_{a_n} = I_a[/spoiler]. But this means that <span class="math">I_a \cap I_{a_1} \cap I_{a_2} \dots \cap I_{a_n} = \emptyset[/spoiler], ...
No. In the first equation the set-relation is wrong, fix it to <span class="math">G_{a_1} \cup G_{a_2} \dots \cup G_{a_n} \supset I_a[/spoiler]. Next, by means of taking complements (which I think is what you intended) this is equivalent to <span class="math">I_a \cap I_{a_1} \cap I_{a_2} \dots \cap I_{a_n} \subset G_a[/spoiler].

>which is a contradiction.
Besides from presumably being miscalculated this does not contradict your assumption which reads <span class="math">\bigcap_{k \in N} I_k = \emptyset[/spoiler] and actually is entailed by your last formula of <span class="math">I_a \cap I_{a_1} \cap I_{a_2} \dots \cap I_{a_n} = \emptyset[/spoiler].

>>6213507
Limit exists because the sequences are monotone and bounded and in R. He said that.

>>6213386
There's also probably several holes that create a nontrivial 2nd homology group

>>6213523
>trivialities I already know
then why did you want to borrow it faggot? If you already know everything in it, why don't you lucidly explain the proof of Urysohn's lemma from memory, or perhaps Tychonoff?

>>6213442
>Take the sequence <span class="math">x_n = \frac{a_n+b_n}{2}[/spoiler] the limit will be an element from <span class="math">\bigcap_{n \in N} I_n[/spoiler] because it is in every I_k .

That's a claim, not a proof. Also notice the fact that all <span class="math">I_k[/spoiler] are closed hasn't been used (and for open intervals, the theorem doesn't hold).

>>6213779
>memorizing proofs

And now we know why you're never gonna get far in math. You treat math like biology.

>>6212154
woah.... this thread... so much retardation, like always...

http://personal.bgsu.edu/~carother/cantor/Nested.html

>>6213860
That is a claim, but it's easy to prove using that all <span class="math">I_k[\math] are closed.

>>6213650
By making those corrections the proof works, though.
Didn't pay enough attention writing it, I started thinking of proving it by contradiction but it's easier to do it directly.
Suppose <span class="math"> I_k [\math] is not empty. Then the last sentence makes sense.[/spoiler][/spoiler]

>>6213860
That is a claim, but it's easy to prove using that all <span class="math"> I_k [/spoiler] are closed.

>>6213650
With those corrections it works, though.
It contradicts the fact that every <span class="math"> I_k [/spoiler] is non-empty and <span class="math"> I_k \subset I_m [/spoiler] if <span class="math"> k > m [/spoiler], so every finite intersection is non-empty.

It may be an incomplete proof, but I don't think it's incorrect.

>>6213865
Have fun getting into any PhD program ever without memorizing tons of bullshit proofs.

>>6213961
Thank you, I will. Do you in your underaged naivety actually believe they are gonna ask a PhD candidate to mindlessly recite arbitrary proofs?

>>6213961
>Memorizing tons of bullshit proofs
More like understanding several important proofs and having the understanding to produce bullshit ones.