#help analysis induction question

106 messages · Page 1 of 1 (latest)

humble kernel
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I figured I could use analysis since I am not saying anything about the elements in the non empty set, just the fact that it is TRUE that it is non empty?

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If I cannot use induction here, please let me know why, thanks.

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I was doing induction on the amount of elements intersected btw

hazy star
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This step doesn't make sense to me

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the approach is to just prove that

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$I_{m+1} \cap \bigcap_{n=1}^{m} I_n$ is non-empty

crude rootBOT
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Magma (N, ^)#Magma4Manager

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hazy star
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For induction

humble kernel
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like from interval m+1 to interval 0

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it's in each previous interval by the fact that they're nested, hence it's in the intersection

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that's all I'm saying by that

hazy star
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hmm wait

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OH

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I see

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I misinterpreted the problem

humble kernel
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I have found another approach that the book intends me to use but I am just wondering if this proof is also valid

hazy star
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ok this proof feels hand-wavy

humble kernel
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how so?

hazy star
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It is pretty much definitional that $a_{m+1}$ is in $I_n$, if that directly implies that $\bigcup_{n=1}^{\infty} I_n$ was non-empty, this wouldn't be a problem

crude rootBOT
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Magma (N, ^)#Magma4Manager

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hazy star
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nv

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m

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it works

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this proof is functional

humble kernel
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Mmm thanks a lot

restive coyote
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i still have some concerns about the proof
starting with N and removing the lowest integer one by one leaves it having elements in it but after infinity, it is the empty set

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but what do i know

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that's not to contest the statement is true, btw
just the general idea

humble kernel
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take N_i to be the set of natural numbers with all natural numbers up to "i" being removed
for i=0, 1 is in the set.
for i=m+1, m+1+1 = m+2 is in the set.

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It looks weird and seems stupid but I'm pretty sure it's how induction works

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but I do see where you're coming from

restive coyote
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the issue is that at every stage, the set is nonempty

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but the limit is empty

humble kernel
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Yeah that makes sense

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but I think induction is just weird like that lol

restive coyote
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wdym

robust sequoia
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You can’t really use induction for this

humble kernel
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how come though

restive coyote
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you can prove that the intersection up to some finite value is nonempty

robust sequoia
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Induction works for finite values

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There are many examples that have properties that hold for finite values but not infinite ones

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Like just because we have a sequence {a}where each a_n is rational does not mean it’s limit is

restive coyote
humble kernel
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mm ok

robust sequoia
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In fact I think this statement is false if we replaced closed sequence with open ones

velvet stone
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to see why it doesnt work, consider nested open intervals. then it doesnt work

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oop beat me to it

restive coyote
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but could it actually be empty

humble kernel
velvet stone
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try using bolzano weierstrass

humble kernel
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I know anotther method

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I just wanted to know if this worked but

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I see why it does not now

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Thanks

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so induction can only be used to prove "forall" values N not "for infinity"

velvet stone
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yes

robust sequoia
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Yeah

velvet stone
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but for some arguments in analysis you can indeed 'pass to the limit'

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so it does sometimes work

humble kernel
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actually

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I have a way to do it without induction but

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I fee like induction would account for open intervals @velvet stone ?

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or you could make it work

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uhh ill write it here later

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I dont think that induction works for this anymore

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but I dont really understand why

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or I do but it's really hand-wavy and unsatisfying

velvet stone
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your induction argument never used the fact the intervals are closed, so it should technically also work for open

humble kernel
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I would like to know exactly why I cant use induction here and in which cases I can

velvet stone
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but its not true for open intervals

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so the argument cant be right

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that was the point i think

humble kernel
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my argument relied on the fact that they were closed though since it literally used the end point as the point included in each interval

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you couldn't use my argument for the open case as either endpoint could be equal I guess

velvet stone
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but if you choose an interior pt and try and use the same argument it fails

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u r essentially saying finite intersection nonempty means infinite intersection nonempty right

humble kernel
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I wrongly assumed that's how induction worked

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like it meant it counted "forall natural numbers" hence it counted for infinity

humble kernel
velvet stone
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no

humble kernel
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oh wait

velvet stone
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(0, 1/n)

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infinite intersection over the ns is empty

humble kernel
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oh interesting I didn't know this

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btw I'm like only starting analysis now

velvet stone
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but any finite intersection is nonempty

humble kernel
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ahh okay interesting thanks

velvet stone
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ig the reason why is because being in many sets does not mean you are in all of them

humble kernel
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eh ok sure thing

velvet stone
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just because i can find someone that belongs to group 1, 2, 3... n for any n doesn't mean i can find someone who is in all the groups at once

humble kernel
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I see

hazy star
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Here is a good example

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$\frac{1}{n} > 0$ for all $n > 1$

crude rootBOT
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Magma (N, ^)#Magma4Manager

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hazy star
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however $\lim_{n \to \infty} \frac{1}{n} = 0$

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as a better example

crude rootBOT
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Magma (N, ^)#Magma4Manager

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hazy star
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$\prod_{n=1}^{k} n$ is finite for all $k$

crude rootBOT
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Magma (N, ^)#Magma4Manager

995223500686118942.png
hazy star
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However $\prod_{n=1}^{\infty} n$ is not

crude rootBOT
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Magma (N, ^)#Magma4Manager

995223500686118942.png