1,452 reputation
317
bio website kmlinux.fjfi.cvut.cz/…
location Prague, Czech Republic
age 27
visits member for 2 years
seen 17 hours ago

Math PhD student at Czech Technical University in Prague and at LIAFA in Paris. At the same time, I'm a typesetter (and partly the copy editor) of one scientific journal (done in LaTeX, of course).

Code licence details applicable on my posts on TeX.SX.

My favourite queries on SE's Data Explorer


1d
awarded  Curious
2d
accepted Is this enough to prove a homeomorphism? — inverse on a dense subset
2d
comment Is this enough to prove a homeomorphism? — inverse on a dense subset
Yeah, I just realized that the problem is eventually simple, and all that needs to be said is that continuous image of a compact is a compact. Thanks for your help anyways for sure!
2d
comment Is this enough to prove a homeomorphism? — inverse on a dense subset
@JohnZHANG Sorry I was not clear. I do have a proof that in my particular case, the 3rd item is true.
2d
asked Is this enough to prove a homeomorphism? — inverse on a dense subset
2d
awarded  Quorum
Oct
19
comment Questions related to maximal ideals
Well, I have seen a definition of a unity in a ring as a non-zero element, which means that the zero ring is then a ring without unity. I know this is strange, but it is exactly for the reason that zero ring is too weird to be a unit ring.
Oct
7
comment How do I find the following definite integral?
Mathematica forever? Not quite in this case :p
Oct
6
comment Fundamental Theorem of Calculus application
It would be wise not to use $x$ for two different variables, it is very confusing.
Oct
3
comment If $A$ is dense in $S$ and $S$ is dense in $T$ , then $A$ is dense in $T$
@KyleStrand Because \bar goes before A, not after that. There's actually a red nothing after the 2nd \bar showing that an argument is missing. The correct thingy is: $\bar S \subseteq \bar A$. And I get your point. Just somehow, you can change balls to "bubbles" (arbitrarily shaped open sets) and it's the same, still keeping the idea behind. And you can even draw it like that without any need of letters -- isn't that cool? :)
Oct
3
comment If $A$ is dense in $S$ and $S$ is dense in $T$ , then $A$ is dense in $T$
@KyleStrand The point is that you first derive that $S\subset \bar A$ implies $\bar S\subset \bar A$, which is a nice, simple and completely general statement, and is somehow the core of the proof. And now, you only use this very general statement, rather than fiddling with complicated stuff as $\epsilon-d(t,s)$, from which, no true idea can be observed.
Oct
2
comment If $A$ is dense in $S$ and $S$ is dense in $T$ , then $A$ is dense in $T$
@KyleStrand I don't like this because it assumes a metrizable space. And while direct and "elementary", it doesn't catch the points at all, IMHO.
Oct
1
awarded  Yearling
Sep
30
awarded  Explainer
Sep
29
comment Is the perimeter of a nested convex set smaller than the containing set's?
@Martín-BlasPérezPinilla Damn, my fault. Sorry for that.
Sep
20
comment Proof my by mathematical induction $\sum_{i=1}^{n} \frac{(-1)^{i-1}}{i} > 0 $
Hello, is really your summand independent of $i$? Because currently the sum evaluates to $(-1)^{n-1}$, which is not always positive...
Sep
19
comment Are $10\times 10$ matrices spanned by powers of a single matrix?
+1 Indeed a very nice proof by contradiction.
Sep
19
answered If A is a matrix, what does A' mean?
Sep
5
comment Can an observed event in fact be of zero probability?
@Did however, the question was on iid $[0,1]$.
Sep
5
comment Can an observed event in fact be of zero probability?
@Aahz No, the fact that you observed it once doesn't mean you'll observe it again later.