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1d
comment Closed subsets of $\mathbb{C}^*$ proper for multiplication
Yeah. I edited and added an overline. But I'm starting to think that you can never have $0\in\overline{S_1}$: $S_1$ is closed so if it has a sequence converging to $0$, it contains $0$ but it can't because it's a subset of $\mathbb C^*$. And so $\overline{S_1}$ can't contain $0$...
1d
comment Closed subsets of $\mathbb{C}^*$ proper for multiplication
Unless I'm mistaken, $(0,\infty)\in \overline{S_1}\times \overline{S_2}$ means exactly $0\in S_1$ and $S_2$ is not bounded. And $S_1\times S_2$ being proper means that for every $r\ge 0$, $S_1\cap \left\{\frac{r}{z}\mid z\in S_2\right\}$ is compact. So your theorem says that if $0\in \overline{S_1\cap \left\{\frac{1}{z}\mid z\in \overline{S_2}\right\}}$, then it's not compact... Whereas it probably should be $\infty\in \overline{S_1\cap \left\{\frac{1}{z}\mid z\in \overline{S_2}\right\}}$. Are you sure it's not $(\infty,0)$ instead of $(0,\infty)$?
Feb
10
comment Test if a number is in ${\mathbb R}$
$x\in \Bbb N\iff x\times x = x + \dots + x$ (for some size of $\dots$).
Nov
22
comment Is $f:\Bbb Q\rightarrow \Bbb R$ continuous?
@ManolisLyviakis Yes. The only problems are with neighbourhoods of $\pm\sqrt{2}$ but since your sequence can't get arbitrarily close to $\pm\sqrt{2}$, by taking a small enough neighbourhood of $\pm\sqrt{2}$, you can make it so that all your sequence is outside of it, and so you avoid the "problem".
Nov
22
comment Is $f:\Bbb Q\rightarrow \Bbb R$ continuous?
The only "problems" are at $\pm\sqrt{2}$ (because everywhere else, you can find a neighbourhood where your function is constant so your function is continuous). Now, let's take a sequence $(u_n)_n$ that converges in $\mathbb{Q}$. To prove that $f$ is continuous, you just need to show that $f(u_n)$ converges (for $u$ arbitrary). But since $u_n$ converges in $\mathbb{Q}$, it can't converge to $\pm\sqrt{2}$ so you're "far enough" from the "problems".
Nov
18
comment why is automaton with a queue is more powerful than an automaton with a stack?
The problem is the one that prevents ${a^nb^nc^n}$ from being context-free: You can only use stored information (the number of $a$) once (to count the $b$s and so then you can't count $c$s). Whereas with a queue, you can use it and immediately put at back in the queue to not forgetting it, while still being able to read the rest of the information.
Nov
12
comment Let X be an infinite set with a topology T, such that every infinite subset of X is closed. Prove that T is the discrete topology.
@zermelovac $T$ is the set of open sets. Since $\{x\}$ is finite and $X$ is infinite, $X\setminus \{x\}$ is infinite and is therefore closed. And saying that $X\setminus \{x\}$ is closed is the same thing as saying that $\{x\}$ is open. So $\{x\}\in T$.
Oct
4
comment What is the property of the eigenvalues of this matrix?
$A^*A=(I-A)A=A(I-A)=AA^*$ so $A$ is normal and you can diagonalize it.
Sep
25
comment Order of operations in polynomial with exponent
You can also factor a $n^2$ out to further simplify.
Sep
25
revised Order of operations in polynomial with exponent
added 2 characters in body
Sep
25
comment Order of operations in polynomial with exponent
Yes. And you should use latex to write it. Just add $ around your math things.
Aug
10
comment Matrix product notation
$\vec{\times}$? And why did you add the tag infinite-product?
Aug
10
comment How do I show that $ \sin x, \cos x$ really are in $ [-1,1]$ using series notion?
@MichaelHardy : If $x$ is real, then $\sin x$ and $\cos x$ are infinite sums of real numbers so they can only converge to a real number. And $(\cos x)^2+(\sin x)^2=1$ can either be derived as in the other answer by differentiation, or you can just compute the series as Cauchy products (which is basically brute force and I thought would be obvious to anyone familiar with series).
Aug
9
answered How do I show that $ \sin x, \cos x$ really are in $ [-1,1]$ using series notion?
Aug
4
comment Degree on Galois theory
And $\left[\Bbb Q(\sqrt[p]{2},\sqrt[q]{2}):\Bbb Q(\sqrt[p]{2})\right]=\cfrac{\left[\Bbb Q(\sqrt[p]{2},\sqrt[q]{2}):\Bbb Q\right]}{\left[\Bbb Q(\sqrt[p]{2}):\Bbb Q\right]}$
Aug
4
comment Degree on Galois theory
Well there's the case where $p=q$ and the case where $p$ and $q$ are coprime.
Jul
15
comment What is the relationship between the concept of a square root and a number's prime factorization?
If $n=2^{i_1}3^{i_2}5^{i_3}\dots$, then $\sqrt{n}=2^{i_1/2}3^{i_2/2}5^{i_3/2}\dots$ so for large numbers that are square, you can compute the factorisation of its squareroot and then double all exponents.
Jun
29
comment Satisfiable formula only over even structares
Hint: If your domain has an even cardinal, you can cut it in two parts (meaning that each element is in one of the parts and no element is in both parts) of equal size (meaning that you have a bijection between the two parts.
Jun
28
comment Need hint to solve a nasty integral.
Can't you just differentiate both sides?
Jun
24
comment Prove that set is bounded but has no max/min
$\frac{s+2}{2}$ is the middle of the segment $[s,2]$. You could take anything between $s$ and $2$ to find a contradiction but the middle is kind of the first thing you think of.