Can anyone help me with these questions? I've been trying, but i couldn't manage to solve them yet

...

sage

>these kind of questions in April

Must be a brainlet CS major in an ""advanced"" "calculus" class

3. If i is in A, let (a_n)= i,i,i,i,i....
If i is not in A, then it is a limit point of A. Therefore the intersection of A and the ball around i of size 1/n is nonempty for all n. Using the axiom of choice, pick a random element from each ball_1/n for a_n. QED.

sigh, I will help but only because I am intro to anal aswell so these problems actually make me think.

3) The infimum is the biggest number that is smaller than or equal to all members of the set so suppose there is no such sequence. In other words, for all possible sequences in A, there exists an epsilon larger than 0 such that the elements of the sequence never get closer to i than by 0.

We can prove that this epsilon cannot get arbitrarily small because if it can then this is the same as the sequence converging to i, and we are assuming that no such sequence does that.

So it must be that there is a minimum epsilon such that the distance between i and an arbitrary element of A is at least epsilon. Then i + epsilon/2 is a number bigger than i such that all elements of A are bigger than it. This contradicts i being an infimum.

4) Literally use the definition of limit man. This is a robot-problem. You could do this simply by blindly following steps, they even gave you the hint.

>then it is a limit point of A
>assuming an analysis course that is just covering convergent sequences always defined what limit points are

>and the ball around
>assuming that an analysis course barely covering sequences already defined balls
>Also, being such a pretintious dickhead that you can't simply call them intervals even when you are working in the number line. It's a ball. A BALL. Oh look at me, I am Mr. Top Topologist in here. Look at me and suck my dick. Look at my pretentious vocabulary.

>Using the axiom of choice
>Assuming that they've gone over the axiomatization of set theory
>Assuming that the professor allows you to assume the axiom of choice
>Assuming that the professor assumes that you assume that he assumes that you can use the axiom of choice

You know [math] (x_n -x) \to 0[/math]. And, using the hint, since [math] (x_n -x) = (\sqrt[3]{x_n}^3 - \sqrt[3]{x}^3) = (\sqrt[3]{x_n} - \sqrt[3]{x})(\sqrt[3]{x_n}^2 + \sqrt[3]{x_n\cdot x}^2 + \sqrt[3]{x}^2)[/math] and [math] (\sqrt[3]{x_n}^2 + \sqrt[3]{x_n\cdot x}^2 + \sqrt[3]{x}^2) > 0[/math] you can deduce [math] (\sqrt[3]{x_n} - \sqrt[3]{x}) \to 0[/math]. Now you just have to translate it to "[math]\varepsilon[/math]-parlance".

>using the order limit theorem
>when you don't even need to

So this is what a university-graduated brainlet looks like?

>look at me, I just proved pythagoras theorem using the axiom of choice and assuming the Riemann Hypothesis

lim (x_n^1/3) = (lim x_n)^1/3 = x^1/3, because x^1/3 is continuos
What do you need the hint for?

>because x^1/3 is continuos

>USING CONTINUITY TO PROVE SOMETHING THAT COULD BE DONE SIMPLY WITH THE DEFINITION OF CONVERGENCE.

>USING CONTINUITY WHEN IT IS EASY TO SEE THAT OP'S ANAL COURSE ISN'T YET IN CONTINUITY

We got some top tier dick jerkers up in here. Some real prestigious TOP Analysis PhDs up in here, thirsty to show how great they are at nuking ants.

>t. eternal virgin

Is this a meme

Is this also a meme?

Maybe I can piggyback a question of mine onto this thread.

If I have a function [math] f [/math], holomorphic on an open set [math] G \subseteq \mathbb{C} [/math], with a root [math] z_0 \in G [/math] of order [math] m [/math].
How can I show that [math] f^{1/m} [/math] has a branch in some open disk centered at [math] z_0 [/math].

I know that if [math] f [/math] has a root [math] z_0 [/math] of order [math] m [/math] I can factor it into the form [math] f(z)=(z-z_0)^m g(z) [/math]
with [math] g(z)=\left\{\begin{matrix}
(z-z_0)^{-m}f(z), z\neq z_0\\
\frac{{}f^{(m)}(z_0)}{m!},z=z_0
\end{matrix}\right. [/math]

I know this reduces to showing that [math] (z-z_0) \sqrt[m]{g(z)} [/math] has a branch in the desired open disk.
I'm somewhat stuck on how to go about this so I'd appreciate a nudge in the right direction.

Since g doesn't vanish at z_0, you can find a small disk D around z_0 such that g(D) lies inside some [math]\mathbb C \setminus e^{i \theta} \mathbb R_{\ge 0}[/math]. Define a log on this domain (which can be done) and set [math]\sqrt{m}{g(z)} = \exp{\frac{1}{m}\log(g(z))}[/math]

shit, I meant [math]\sqrt[m]{g(z)} := \exp\left(\frac{1}{m}\log(g(z))\right)[/math]

>g doesn't vanish at z_0
I understood that, however the [math] (z-z_0) [/math] term still vanishes at [math] z_0 [/math] which I thought was the problem with finding the branch.
I don't have a lot of experience with the general idea behind finding a branch for a function I guess. Maybe could you dumb it down a little more for me :( thanks for your help though user

Not a meme.

From the questions given, you can tell this is a first week analysis question (maybe even first class). So you can't assume ANYTHING.

You can't assume that you know the topology of R. (What one of you was doing). And you can't assume continuity because even if you proved the function was continuous, the professor still hasn't taught that so he would not accept that reasoning.

Come on, you know this.

>>Using the axiom of choice

AoC is done in the intro to proofs course before analysis. Actually since this is April, this homework question is surely coming from a proof course that moved into baby's first analysis topics for practice.

>>Assuming that they've gone over the axiomatization of set theory

Everyone mentions the AoC in naive set theory.

>>Assuming that the professor allows you to assume the axiom of choice

Who doesn't?

Gr8 b8

>AoC is done in the intro to proofs course before analysis. Actually since this is April, this homework question is surely coming from a proof course that moved into baby's first analysis topics for practice.

Lets be real, probably not. Sure, it *COULD* happen but when has an intro to proofs course done that? If it not number theory then intro to proofs never gets into the specifics because then it is not really an intro to proofs class anymore.

>Everyone mentions the AoC in naive set theory.

Yeah but do you get told to use it? Never, because you don't need it. No one even needs the AoC until you go really deep into a specialty, or you go really meta and undergrad is neither specialty not meta.

>Who doesn't?
Why. would. them.

Proving something as trivial as this with the AoC just outs you for a brainlet.

>[math]x \in \mathbb{R} [/math]
You should probably drop out of this class and look for a better lecturer. It is a well-known fact (even if it is suppressed by the established "mathematicians") that the "real numbers" are an utterly incoherent concept, like every other construction depending on the notion of "infinity".
People will tell you otherwise but it's just because they don't want to lose their cushy high-paying jobs.

BUT WILDBERGER
EVERYTHING THAT IS TRUE ABOUT THE REAL NUMBERS IS APPROXIMATELY TRUE FOR THE RATIONAL NUMBERS (THE NUMBERS OF THE REAL WORLD)

IF IT WEREN'T FOR THESE APPROXIMATIONS WE ALLOW OURSELVES TO ASSUME WE WOULD NOT HAVE ANY KIND OF TECHNOLOGY OR ADVANCEMENT. YOU WOULD NEED RIDICULOUSLY SPECIFIC ALMOST INFINITE ARGUMENTS ABOUT SEQUENCES OF RATIONALS TO ACHIEVE WHAT YOU CAN WITH REALS

SO WHAT WILL YOU DO? ACCEPT A LITTLE OF INCONSISTENCY WITH YOUR AXIOMS OR PERISH LIKE A DOG?

nvm I figured out what you mean thanks a bunch user

You use choice when you get to cardinality. Prove that if you have 2 onto functions f: A->>B and g: B->>A, that A and B are equinumerous.

It is not fucking AC, it is countable choice and I'm sorry but you're going to miss out on a lot of analysis if you don't allow countable choice because proofs of nice sequential criteria for continuity/closedness/compactness/whatever in metric spaces (ie. the whole reason we like metric spaces) are going to rely on countable or even dependent choice.
Most profs don't even mention it because it's so obvious you can't do anything without it (unless you contort yourself way beyond what is reasonable for an intro class).

Note |y-x| < delta ==> y-x y < delta+x.
Thus, for small enough delta, |y-x| = |y^1/3 ^3 - x^1/3 ^3|
= |y^1/3 -x^1/3| ( y^2/3 + y^1/3 x^1/3 + x^2/3 )
< delta ( (delta+x)^2/3 + (delta+x)^1/3 x^1/3 + x^2/3 ) < epsilon,
since the limit is zero as delta -> 0.

>Every uni is like mine.
Now this is autism.