help

[mental chasm]

i tried this using the the proof on the right paper, i got the feedback, "you need to suppose "E is not closed", not "E complement is not closed", to prove the statement by contradiction.'' , so how can i change this then?

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[neat tangle]

change what?

[mental chasm]

my proof

[neat tangle]

first of all if you work in R^3 as the main metric space, then you might want to assume A complement is not open instead because a set is closed iff its complement is open

[regal cave]

Suppose E is not closed. Then E complement is not open.
Let x be a point in E complement. Prove that it has an open neighbourhood entirely in E complement. Hence E complement is open; contradiction arrived.

[neat tangle]

because being open doesn't imply it's not closed!

[lapis cipher]

what definition of open or closed do you have?

typically closed means complement is open

are you allowed to use sequences?

$$ E\ni (x_1^n,x_2^n,x_3^n) \xrightarrow[n\to\infty]{}(x,y,z) \in \mathbb R^3 $$

[livid gull]

aL

[lapis cipher]

this implies x,y,z in E

that's the easiest approach I see

[regal cave]

Wouldn't that be a direct proof

[lapis cipher]

" i got the feedback, "you need to suppose "E is not closed", not "E complement is not closed""

I take it this means IF you want to prove by contradiction, then so and so

but I didn't read it was mandatory to prove it by contradiction

otherwise you'd have to do epsilon stuff.

Suppose E is not closed. Hence its complement is not open. Hence, there exists a point in R^3 \ E, that is not an interior point. So any epsilon ball around it would have to intersect with E

[neat tangle]

following up what aL said, it's provable than $(x_n, y_n, z_n) \to (x, y, z)$ if a only if $x_n\to x$, $y_n\to y$ and $z_n \to z$ under the euclidean metric As $\qty{x_n} \subseteq \mathbb{R}{\geq 0}$ which is closed we have $x\in \mathbb{R}$ and the same follows for the rest and we get $(x, y, z) \in \mathbb{R}^3{\text{non negative coordinates}}$

[livid gull]

Coffey

[neat tangle]

(just elaborating for the guy he might find it interesting)

[neat tangle]

and this logic can be used to prove that the finite Cartesian product of closed sets is closed too

Now if it becomes infinite that's where things become funny

[lapis cipher]

I would assume sequences are allowed because at this stage we have covered sequence based topology

but who knows..

[neat tangle]

sometimes people don't like the usage of sequences

high chance of them not being allowed in proofs

[mental chasm]

ok but then

[mental chasm]

if i would do this, how do i prove that is has an open neighbourhood entirely in E complement

[regal cave]

Since x is in E complement, at least one of its coordinates is negative; call it x_0

Then the open ball of radius |x_0|/2 = -x_0/2 lies entirely in E complement

[mental chasm]

so if i would write : Suppose E is not closed. Then E complement is not open.
Let x be a point in E complement. Since x is in E complement, at least one of its coordinates is negative; call it x_0
Then the open ball of radius |x_0|/2 = -x_0/2 lies entirely in E complement. Hence E complement is open; contradiction arrived

that would be correct

[lapis cipher]

yes, because x is arbitrary

but at this point it's silly to formulate it as a proof by contradiction

you showed the complement is open, hence by definition E is closed

@mental chasm

[regal cave]

True

Either way the proofs are epsilons away from their direct proof counterparts

[mental chasm]

so then how would you formulate this assignment?

[lapis cipher]

To show E is closed it suffices to show its complement is open.Let x be a point in E complement. Since x is ..
qed

[mental chasm]

thank you

[neat tangle]

+close