polite proofs

@MordeusMorgenstern

Can I ask what year you are in?

I've also accepted your answer

Thank you

And I'll waste less time doing 'lazy' work where I just sit there not doing too much

I think the biggest thing that helps is switching to another thing for a while, like topology. Then I'll be more motivated coming back to do more analysis

But it's kind of difficult to think about things deeply without spending an absurd amount of time on them, but it's essentially required

I try

Hmm, okay then

Why not? I'm very far from learning multivariable analysis rigorously

I just wanna know multivariable calculus for the sake of knowing it

Okay, I'll have a look at it, thanks. Did you want to post a response to the question so I can accept it?

Haha, well, I just don't want to get too hung up on doing topology since I really want to finish my current plans as soon as possible

Strange I've previously heard good things about Munkres' book

Hmm, okay, I'll look into that book then

Or why do you say it's bad

Are you talking about the later chapters?

By analysis I mean more generalized analysis, since Spivak doesn't really do topology

I plan to get back to it after I finish Spivak. Current plan is Munkres initial chapters -> Back to Spivak -> Computational multivariable calculus -> Linear Algebra (rigorously) -> Analysis, probably by Rudin or something

I'm in my early 20's, and yes

So now I'm doing Spivak and I'll do the first some chapters of Munkres until I can't anymore

But I actually think linear algebra is often more straight forward at that level

Actually no, I did read the first few chapters of Axler before I quit

Then yes, but basically I didn't learn anything for over a year (to learn mechanics and chemistry) and then recently restarted reading my first rigorous texts

The last option, but I don't know how recently you would consider recently

Well that depends on your definition of starting out

:) I am glad

Would that work?

Let $y \in B_0$. By surjectivity, then there is some $a \in f^{-1}[B_0]$ such that $f(a) = y$. Therefore, $y \in f[f^{-1}[B_0]]$ which implies the desired result

Okay uh

I think we would reach a contradiction if f^{-1}(B_0) is empty?

Right, what's wrong with this? I am just applying the definitions and it should be true unless B_0 is empty

But what about my argument currently bothers you?

Hello Captain

@MordeusMorgenstern I don't need it, because we're not talking about $y$ just yet; we don't need surjectivity because we implicitly assume the sets are nonempty.

@MordeusMorgenstern Basically, I wanted to first state why there exists such an element in those sets, and then I apply surjectivity to get that it 'reaches' $y$.

@MordeusMorgenstern What about the rest of the proof?

I understand what you mean, yes that's definitely not the intended interpretation. I will edit my post.

@MordeusMorgenstern Hi, I agree with what you said, including that your suggestion is less confusing, however why is it wrong?

I guess since $L(f,P) = k - U(g,\psi(P))$ for any partition $P$ of $[a,b]$...?

@TedShifrin Why does one $k$ work in his case?

The reasoning just seems really handwavy..

I don't know... seems dodgy

Okay, well, we know that $U(f,P) = k - L(g,\psi(P))$. So then over all partitions $P$ of $[a,b]$: $\sup U(f,P) = \sup \{ k - L(g,\psi(P)) \}$. The impression inside the braces is maximized when $L(g,\psi(P))$ is minimized......

@TedShifrin Wait, so how do we justify it then?

Choosing $P$ locks in $P'$

Right

I just feel like he does something very similar.

Hmm

But that's essentially what Spivak does in the proof of 13-21 which I just realized