Mathematics - 2024-12-13

Jakobian

00:08

@sku $1$ is special because it doesn't converge uniformly and the problematic point is precisely $x = 1$

oh. I guess you mean this specific limit?

psie

09:05

@Jakobian ok 👍 you said pick your favorite metric on the product $X\times X$. If $X=\overline{\mathbb R}$ and I've chosen $d(x,y)=|\arctan x-\arctan y|$, then I could simply choose $\max(d,d)$, or?

psie

09:23

Here's my attempt, however, without using sequences this time.

Attempt Choose $(x,y)\in X\times X\setminus\{(a,a): a\in X\}$. Let $c=d_X(x,y)/2>0.$ To show $B((x,y),c)\subset X\times X\setminus\{(a,a):a\in X\}$, suppose for the contrary that $(h,h)\in B((x,y),c)$.

Then \begin{align*}d_X(x,y)&\le d_X(x,h)+d_X(y,h)\quad\text{(triangle inequality)}\\ &\leq 2\cdot \max\{d_X(x,h),d_X(y,h)\}\quad\text{(definition of max)}\\ &= 2\cdot d((x,y),(h,h))\quad\text{(definition of product metric)}\\ &< 2\cdot c\quad\text{(assumption on $h$)}\\ &= d_X(x,y)\quad\text{(definition of $c$)}\end{align*}

In other words $d_X(x,y) < d_X(x,y)$, contradiction.

If this is OK, then I'm more than happy to not bother about this anymore :)

Jakobian

Did you check if its ok yourself already?

psie

@Jakobian well, there's one small detail that I'm a tiny bit unsure about. Let $A,B,C$ be sets, $C\subset B$. Is the negation of $A\subset B\setminus C$ that there exists a point $x\in A\cap C$?

Jakobian

10:57

@psie what do you think

psie

11:12

@Jakobian since I used it in the proof above, obviously I think it holds (with the additional condition that $B$ is the universe). However, a confirmation would be nice.

ILikeMathematics

12:14

Does anyone have a solution to the pretty well known result that no continous function over [0, 1] takes on every value n times?

Does it 'easily' extend from n = 2 to generally n>= 2?

Simd

12:37

This is about n by n binary matrices. Call a pair of rows unfriendly if the Hamming distance between the two rows is larger than the number ones they have in common positions. Let us call a matrix unfriendly if all pairs of rows are unfriendly.

The identity matrix is, for example, unfriendly and the all ones matrix is friendly.

What is the largest number ones an n by n unfriendly binary matrix can have?

Thorgott

13:21

@ILikeMathematics I've not heard of this well-known result, but I suppose it's not too difficult? Look at the $n$ maxima, take a slightly smaller value and use the IVT to see that it has to be attained at least $n+1$ times.

Jakobian

@psie do you think it holds or did you convince yourself it holds

this is too basic for me to try and convince you if it holds or not

you should do such things yourself

ILikeMathematics

13:37

Proved it, thanks

ILikeMathematics

14:31

@Thorgott Oh, and btw

$$f(x) = \frac{x+1}{x-5} \text{ on $D= [1, 5)$}$$ is continuous, right?

\begin{align*}
|f(x) - f(y)| &= \left|\frac{x+1}{x-5} - \frac{y+1}{y-5}\right| = \left|\frac{(x+1)(y-5) - (y+1)(x-5)}{(x-5)(y-5)}\right| \\&= \left|\frac{xy - 5x + y - 5 -(yx - 5y + x - 5)}{(x-5)(y-5)}\right| = \left|\frac{-6x + 6y}{(x-5)(y-5)}\right| \\&= 6 \cdot \frac{|x - y|}{|x-5||y - 5|} < 6\delta \frac{1}{|x-5||y - 5|}
\end{align*}

Thorgott

yes, a rational function is continuous everywhere where it is defined

ILikeMathematics

We can bound $\delta$ to get $-1 < x - y < 1$ and so $-1 -x < - y < 1 - x \iff 1 + x > y > x - 1$, so $x - 4 > y - 5 > x - 6$: $$6\delta \frac{1}{|x-5||y - 5|} < 6 \delta \frac{1}{|x - 5|\max\{|x-6|, |x-4|\}} < \varepsilon,$$ thus $$\delta < \frac16 \varepsilon|x-5|\max\{|x-6|, |x-4|\}.$$

@Thorgott Does this proof look fine to you? I don't like the max in there

Thorgott

I'd recommend against doing this by hand, but also the first inequality involving max seems wrong

ILikeMathematics

@Thorgott Oh?

$x - 6 < y - 5 < x - 4$

I thought we could conclude that, then

Thorgott

14:48

this gives you $|y-5|\le\max\{|x-4|,|x-6|\}$, not the other way round

ILikeMathematics

Oh

Thorgott

:)

ILikeMathematics

@Thorgott Any idea on bounding the denominator then?

We have $$6\delta \frac{1}{|x-5||y - 5|}$$

Thorgott

reverse triangle inequality should probably do it

ILikeMathematics

@Thorgott On what?

Thorgott

14:53

if $y$ is close enough to $x$, it can't be too close to $5$

ILikeMathematics

@Thorgott $||x| - |y|| \leq |x - y|$

If $|x - y| < 1$, then $|x - 5| = |x - y + y - 5| \leq |x - y| + |y - 5| < 1 + |y - 5|$

Do you mean something along these lines?

@Thorgott How do we express that?

Thorgott

with the reverse triangle inequality

ILikeMathematics

$||x - y| - 5| \leq |x - y - 5|$

@Thorgott Like that?

pie

16:03

Do you think Complex analysis by Elias M. Stein would be a good choice for an introduction?

Jakobian

16:19

@ILikeMathematics If $f_1, f_2$ are continuous functions into $\mathbb{R}$, then so are $f_1+f_2, f_1\cdot f_2$ and if $f_1/f_2$ is well-defined then its also continuous

Two obvious facts: the identity function $x\mapsto x$ and the constant functions $x\mapsto a$ are continuous

it follows from this that any polynomial is continuous

and so, any rational function where it is defined

no need for computations - the computations are already done when proving that $f_1+f_2, f_1\cdot f_2$ and $f_1/f_2$ are continuous

so there's no purpose in doing it by not using those theorems - you are doing the same thing either way

if those theorems are not available, then same amount of work will be had by proving them

Soumik Mukherjee

16:33

@pie yes, it has good exercises

psie

20:21

What are some easy ways to show the Borel $\sigma$-algebra of $\mathbb R$ is generated by all compact sets? This is a claim I'm facing. I'm used to seeing this $\sigma$-algebra being generated by the open sets, or various intervals of different forms.

I'm interested in all the compact sets because I think they form a $\pi$-system.

psie

20:33

I found a way I think, but is it trivial that the class of compact sets of $\mathbb R$ is a $\pi$-system?

Thorgott

closed rectangles are compact

psie

yes? I think the collection of compact sets of $\mathbb R$ is a $\pi$-system because compact sets are closed and bounded, and these properties are preserved under finite intersections. Maybe I'm mistaken.

Thorgott

I'm saying that closed rectangles are compact implies the Borel algebra is generated by compact sets

I neither remember nor care what a $\pi$-system so I'm not commenting on that part

Jakobian

21:05

@psie open sets are countable unions of compact sets

@psie yes. Closed under arbitrary intersections

psie

nice, thanks.

@Jakobian are you looking forward to Christmas this year? :)

It's approaching with great speed.

Jakobian

If you take $\mathbb{Q}$ and add two points $x_1, x_2$ such that nbds of them are $\{x_1, x_2\}\cup U$ where $\mathbb{Q}\setminus U$ is compact, then $A_i = \{x_i\}\cup \mathbb{Q}$ are compact but $A_1\cap A_2$ is not

So intersections of compact sets need not be compact

However, intersections of compact closed sets are always compact and closed

For Hausdorff space, a compact set is always closed

@psie It'll probably be a sad Christmas so no

psie

21:22

@Jakobian that's a shame to hear :(

pie

22:34

@SoumikMukherjee It is the second book in "Princeton Lectures in Analysis" and the first one is : Fourier Analysis: An Introduction, should I read this before the complex analysis book?

Xander Henderson

@pie Are you talking about Stein and Shakarchi?

pie

@XanderHenderson yes

Xander Henderson

My feeling is that the books in the series are fairly self contained, but I only encountered them in grad school after taking a lot of real analysis.

So YMMV.

But you don't need Fourier anal to do complex anal.

pie

@XanderHenderson would you recommend this book for an introduction to complex analysis?

Xander Henderson

@pie I haven't read the complex anal book in the series. But I think that both the Fourier analysis and real analysis texts are pretty approachable.

I can't imagine complex being of a significantly different quality.

I will note that I think that the exercises often don't quite match the exposition in the text (the exercises are generally good, but they often seem to be a little bit disconnected---my pet theory is that Stein wrote most of the book, and had his grad student write the exercises).

think_meaning_buildß

22:50

it also greatly depends on what you mean by "an introduction to complex analysis"

pie

@think_meaning_buildß How can there be different meanings to that?

I really don't know how to answer that, or what the answer should be.

think_meaning_buildß

varying levels of sophistication; background knowledge through applications

pie

@think_meaning_buildß Primarily Baby rudin real analysis and Hoffman and Kunze linear algebra

think_meaning_buildß

then you should be ok

Sine of the Time

23:06

@Soumik Russians are accusing Ding of losing on purpose

Xander Henderson

Honestly, Stein and Shakarchi is probably the gentlest introduction to the topics that I have seen. If their complex book is anything like the others, and it goes over your head, you likely need to review something else, and not find a different complex book.

If you've worked through baby Rudin, I would guess that Stein and Shakarchi will be fine.

think_meaning_buildß

23:29

yeah, and if you did it alone you should be applauded

Dimitrios ANAGNOSTOU

Hello to all. What is wrong with my question here? math.stackexchange.com/questions/5011099/… Why the negative voting?

think_meaning_buildß

I only see one negative vote?

Drive-by down voting is not worth questioning anyway.

copper.hat

23:43

introduction just means that. let me introduce you to my pet lion

2

i liked Conway's Functions of one complex variable.

Xander Henderson

@copper.hat That's the one I learned out of. Seemed fine.

copper.hat

while not usually a fan, i liked the first few pages of Rudin's RCA

Xander Henderson

Ha!

copper.hat

@DimitriosANAGNOSTOU are you asking if it has a solution when when the $\det$ is zero?

think_meaning_buildß

@copper.hat does the ~~command~~ request "here kitty kitty kitty" still apply?

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$Mathematics$ Mathematics