norm spaces

[edgy wedge]

can someone please explain to me what it means for these sequences to be a closed subspace?

[junior peak]

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[prime crypt]

closed means complement is open

in metric spaces closedness is convenient to check with limits

e.g for 1. suppose you have a sequence of elements in l ^inf satisfying a_0 = 0 converging to some element in l ^inf, is that element also in this set?

@edgy wedge

as usual, convergence means norm of the difference converges to zero

[edgy wedge]

can you please clarify what sequence of elements means

im still trying to wrap my head around norm spaces

[prime crypt]

you've dealt with sequence of real/complex numbers yes?

[edgy wedge]

just random sequences?

like a_n=1/n, a_n=n^2 etc?

[prime crypt]

for example

now you have a sequence of elements of this set

[edgy wedge]

so we have sequences that are part of l^inf with the initial condition a_0=0?

[prime crypt]

yes, and we assume the sequence converges to some element in l^inf

is that element also necessarily in this set?

[edgy wedge]

l^inf just means sup is a certain number right

like we can take the sup of a sequence

[prime crypt]

that's the norm in l^inf

[edgy wedge]

oh

[prime crypt]

$$ |(x_1,x_2,\ldots)| := \sup |x_i|$$

[red wind]

aL

[edgy wedge]

ye okie

so what does l^inf mean?

sorry i couldnt follow my lecturers explanation so im struggling with bunch of concepts in functional analysis

[prime crypt]

look up the definition

[edgy wedge]

so a sequence in l^inf is just bounded?

[prime crypt]

yes, because otherwise

[prime crypt]

this could be infinite

[edgy wedge]

ye

[prime crypt]

which is no good

[edgy wedge]

wdym in norm of differences?

[prime crypt]

in any normed space

$$ a_n \to a \Leftrightarrow |a_n-a|\to 0 $$

[red wind]

aL

[edgy wedge]

okie

[prime crypt]

now apply it to your problem

[edgy wedge]

there's a value a s.t $\norm{a_n-a}<\epsilon \forall \epsilon>0$

[red wind]

naokye

[edgy wedge]

sooo

we're given a_n is in l^inf

hence msut be bounded

[prime crypt]

firstly what does it mean in l^inf norm terms?

[edgy wedge]

sup(x_n)

wait

sup|x_n|

[prime crypt]

norm of difference

[edgy wedge]

sup|a_n-a|?

[prime crypt]

yes

but more specifically

[edgy wedge]

sup|a_n-a|<e

[prime crypt]

$$ \sup |a^n_i-a_i| $$

[edgy wedge]

for all e>0

[prime crypt]

you don't take supremum of the sequence but its components

[edgy wedge]

that ^n isn't a power right

[prime crypt]

it's an index

[edgy wedge]

okie

so the differenece of different sequences?

[prime crypt]

$$ a^n = (a^n_0, a^n_1,\ldots) $$

[edgy wedge]

so we have a sequence of sequences?

[prime crypt]

correct

[edgy wedge]

damn

this is wild

[prime crypt]

your sequences start with index 0 tho yes?

[edgy wedge]

ye

[red wind]

aL

[edgy wedge]

but we have sequence of sequences but all starts with a_0=0?

[prime crypt]

for each n a^n_0 = 0

by assumption

[edgy wedge]

ye

[prime crypt]

does it follow that a_0 = 0?

[edgy wedge]

wdym in this?

[prime crypt]

that's what you have to check for closedness

is the limit of the sequence a^n also in the set?

[edgy wedge]

i see

[prime crypt]

and what's the condition for a sequence to be in the set..?

[edgy wedge]

what did you mean in this tho?

[edgy wedge]

it converges

and a_0=0

[prime crypt]

not converges

bounded

[edgy wedge]

oh ye

ofc

[prime crypt]

it has to be l^inf and also a_0 = 0

we can assume a in l^inf, so we only need to check a_0 = 0

if this is true, then the set is closed

[edgy wedge]

ohhh right

[prime crypt]

$$ a^n_0 = 0\forall n,\ \sup |a^n_i-a_i| \to 0 \overset{?}\Rightarrow a_0 = 0 $$

[red wind]

aL

[prime crypt]

the answer is an obvious yes

just think about what it means when supremum over all indices converges to 0

[edgy wedge]

indice means every individual sequence right?

[prime crypt]

supremum is taken w.r.t components of the sequence

[edgy wedge]

okie...

[prime crypt]

see this

[edgy wedge]

did we assume that a_n converges?

[prime crypt]

yes

a^n -> a by assumption

that's this part here

[edgy wedge]

okie ye this is decently ocming together now

so for a sequence to be closed, the limit converges to smth in the set?

[prime crypt]

set is closed if and only if it contains all its limit points

that's what we're testing

we take an arbitrary limit point of this subset and check whether it also is contained in the subset

[edgy wedge]

okie

tysm for your help!

ill trye to attempt the otehr two questions now

.close

[winged sequoia]

Unable to parse the channel name

[edgy wedge]

oh

+close