@Jakobian I can't think of anything, but the suggestion about every closed set being the closure of a singleton is very restrictive. it implies every closed subset is irreducible, so the collection of closed subsets is totally ordered. since intersections of closed subsets are closed, it is well-ordered. if your space is T_0, this should yield that it's an ordinal space.
@Jakobian Thank you very much; I will have to start reading on these things seriously, when I find the time. Thus far, i am not studying from a book (on set theory) right now, I am just trying to prove random results I come across, that make use of AC (cuz they seem like cool questions).
Like eg: I came across the theorem $| S \times S |=|S|$ and the theorem $|S|=|\text{ set of all finite subsets of }S|$. There are others, like $|\text{Any linearly independent set}| \leq |\text{Any spanning set}|$ which I showed yesterday, and another one that I had was "$dim(V) \leq dim(\mathscr{L}(V,F))$ with equality iff finite dimension".
Another one I had was every set can be equipped with group structure, havent given this one much thought. These are just random problems that I give a go when I find some free time, or if im really bored with my other studies
there seems to be a million statements equivalent to AC, each in different contexts- analysis, topology, algebra etc
which is also a cool thing ig
ofc now I can see to attack these problems one might need more knowledge
@Pizza I think I've found your mistake. Using the convention $F(f)=\int f(u)\exp(-2\pi i u k ) du$, for the properties of the FT you can compute $\int u^2e^{-|u|} \exp (-2\pi i u k) du$
denote by $a=2\pi i k$, and split the region of integration: $\int_0^{+\infty} u^2 e^{-u}e^{-au} du=\int_0^{+\infty} u^2 e^{(-1-a)u} du=\int_0^{+\infty} u^2 e^{-b u} du =\frac 2{b^3}$
@SineoftheTime using $\frac1{(a+1)^3}+\frac1{(1-a)^3}=-\frac{2(1+3a^2)}{(a^2-1)^3}$ with $a=2\pi i k$, you get $\frac{2(1-12\pi^2 k^2)}{(1+4\pi^2 k^2)^3}$
Multiplying by $2\exp(-10 \pi i k)$, $\mathcal F(f)= \frac{4-48\pi^2 k^2}{(1+4\pi^2 k^2)^3}e^{-10 \pi i k}$
$$\int_{0}^{\infty }\sin(x^{2})dx$$
I'm confused with this integral because the square is on the x, not the whole function. How can I integrate it? Thank you.
I have not done complex analysis (only real analysis as I am a high school student) so how can I evaluate it using elementary functions ...
I've used the Gamma function: $\int_0^{+\infty} u^2 e^{-bu} du\overset{ bu = y}{=} \frac1b \int_0^{+\infty} \frac{y^2}{b^2} e^{-y} dy =\frac1{b^3}\Gamma(3)=\frac 1{b^3}2!$
but when I evaluated the first by parts I then noticed that the second integral was the same only that this changed: instead of having the exponent (1-ik) I had (-1-ik)
I evaluated both because thet had different extremes, but perhaps there are some shortcuts
I'm reading a proof concerning the statement that $L^p(E,\mathcal A,\mu)$ is a Banach space. In proving completeness, there's this one bound I simply do not understand. We have a sequence of functions $(g_n)$ for which $$\|g_{n+1}-g_n\|_p\leq 2^{-n}.\tag1$$Then the author uses the following bound for $m>n$; $$\|g_{m}-g_n\|_p\leq \|g_{n+1}-g_n\|_p+\cdots +\|g_{m}-g_{m-1}\|_p\leq 2^{n-1}.\tag2$$I really don't understand $(2)$, especially the last inequality.
Any clarification would be very appreciated.
Meeh. I have a typo. The $2^{n-1}$ in $(2)$ should be $2^{-n+1}$. Sorry.
Ok, I think I see what's going on. We have that $\|g_m-g_n\|_p\leq \sum_{i=n}^{m-1}2^{-i}$, which in turn is bounded by $\sum_{i=n}^{\infty}2^{-i}=2^{-n+1}$.
I'm reading a proof about the fact that $L^\infty(E,\mathcal A,\mu)$ is a Banach space. I'm paraphrasing. First we establish that $\|\cdot\|_\infty$ is a norm and then we look at a Cauchy sequence $(f_n)$ in $L^\infty(E,\mathcal A,\mu)$. We show that $|f_n(x)-f_m(x)|\leq\|f_n-f_m\|_\infty$ for every $m>n\geq 1$ and $x\in E\setminus N$, where $N$ is null. We realize that $(f_n(x))$ is Cauchy in $\mathbb R$ and converges to $g(x)$ (for $x\in N$, we simply put $g(x)=0$).
Then $g(x)$ is measurable and if we take the limit as $m\to\infty$ in $|f_n(x)-f_m(x)|\leq\|f_n-f_m\|_\infty$, we obtain $$\sup_{x\in E\setminus N}|f_n(x)-f_m(x)|\leq\sup_{m\in\{n+1,n+2,\ldots\}}\|f_n-f_m\|_\infty.\tag{1}$$
I just don't understand how $(1)$ was obtained. Is someone well-versed in this proof? Any clarification would be appreciated.
I'm reading a proof about the fact that $L^\infty(E,\mathcal A,\mu)$ is a Banach space in Le Gall's Measure theory, Probability and Stochastic processes.
I'm paraphrasing the proof. First we establish that $\|\cdot\|_\infty$ is a norm and then we look at a Cauchy sequence $(f_n)$ in $L^\infty(E,\...
I fixed an important error there; $(1)$ above should read $$\sup_{x\in E\setminus N}|f_n(x)-g(x)|\leq\sup_{m\in\{n+1,n+2,\ldots\}}\|f_n-f_m\|_\infty.$$
$$\int_{0}^{\infty }\sin(x^{2})dx$$
I'm confused with this integral because the square is on the x, not the whole function. How can I integrate it? Thank you.
I have not done complex analysis (only real analysis as I am a high school student) so how can I evaluate it using elementary functions ...
