Reputation
Next tag badge:
90/100 score
35/20 answers
Badges
1 17 56
Newest
 Nice Answer
Impact
~201k people reached

2d
comment Homeomorphism of a triangle.
"In your comment about "topologically rigid", you presumably meant "metric space" rather than "topology space", because the notion of "isometry" is defined only in the presence of a metric.
2d
comment Homeomorphism of a triangle.
Another example: Circumscribe a circle around your triangle, and let $D$ be that circle together with its interior. Let $p$ be the homeomorphism from your triangle to $D$ given by radial projection (linear on each radius), and let $r$ be a rotation of $D$ around its center. then $p^{-1}\circ r\circ p$ is a homeomorphism. It's an isometry only when the rotation $r$ is by a multiple of 120 degrees.
2d
comment Homeomorphism of a triangle.
Let the vertices of your triangle be A, B, and C; let P and Q be two distinct points in the interior of the triangle. Define $f$ to (1) map each of $A,B,C$ to itself, (2) map $P$ to $Q$, and (3) be linear on each of the triangles $APB$, $BPC$, and $CPA$. that's a homeomorphism but not an isometry.
2d
comment Homeomorphism of a triangle.
You can't prove that $f$ is an isometry, because that doesn't follow from the given information.
2d
answered Q has no maximal subgroups.
Apr
14
answered Induced homomorphism - Homology
Apr
14
comment Induced homomorphism - Homology
For the quotients $X/H$ and $Y/H'$ to be groups, you need that $H$ and $H'$ are not merely subgroups but normal subgroups of $X$ and $Y$ respectively. That would be automatic if $X$ and $Y$ were abelian groups, but you haven't assumed that. Furthermore, in order to get an induced homomorphism $f^*$, you'd want two additional assumptions, namely that $f$ is a homomorphism (not just a function) and that $f(H)\subseteq H'$.
Apr
14
comment Induced homomorphism - Homology
It's far from clear what you're asking about. Arbitrary functions $f$ don't induce homomorphisms. Even homomorphisms $f:X\to Y$ won't induce homomorphisms $f^*:X/H\to Y/H'$ unless you assume some relationship between $f$, $H$, and $H'$. Furthermore, you never said what sort of algebraic structures you're talking about: groups? modules? rings? or what? Your explanation "$f^*(a)=b$ where $a\in X/H$ and $b\in Y/H$" is just repeating that $f^*:X/H\to Y/H'$ and says nothing about "keep[ing] the structure".
Apr
11
comment is differ between distributive lattice vs semi-lattice on Turing Degrees
As Carl Mummert explained in his answer, $(D,\leq)$ cannot be a distributive lattice because it is not a lattice; it has two elements with no greatest lower bound.
Apr
10
comment is differ between distributive lattice vs semi-lattice on Turing Degrees
From the second sentence of the question, I conjecture that you're conflating the poset of Turing degrees (partially ordered by Turing reducibility) and the lattice of r.e. set (partially ordered by inclusion).
Apr
10
answered Borel sigma-algebra over [0,1]
Apr
9
answered $\forall \alpha \exists \beta: \beta > \alpha$ where $\alpha$ and $\beta$ cardinals
Apr
9
comment Combinatorial homework problem.
What is the effect of any single step on the total number of piles? How many piles are there at the beginning of the process? How many are there at the end? I think all three of these questions are easy to answer, and putting the three answers together will solve the problem.
Apr
4
comment Value of $\int_{-\infty}^{\infty} \delta(t-\pi)\cos(t) \,dt$
I suggest making the substitution $u=t-\pi$ and then using the definition of the $\delta$-function (which tells you the integral of the product of $\delta(u)$ with any smooth function) to get the answer, $-1$.
Apr
4
comment Is every isomorphism of an algebraically closed field onto one of its subfields an automorphism?
If $\overline F$ means the algebraic closure of $F$, and if $\tau:\overline F\to\overline F$ is a homomorphism extending the identity on $F$, then $\tau$ is surjective, because its range is algebraically closed. So such a $\tau$ is an automorphism of $\overline F$. Nevertheless, the answer to your original question remains negative, as Zhen Lin pointed out.
Apr
4
comment Trying to calculate the integral limit $\lim_{n\rightarrow\infty} \int_{-\sqrt n}^{\sqrt n}\left (1 - \frac{x^2}{2n}\right)^ndx$
I guess the question becomes easier if one labels it "freshman calculus" instead of "real analysis".
Apr
4
comment Limit of a sequence - the sign of infinity
Or maybe divide by $n^4$.
Apr
4
comment Contructive implications of Cantor's diagonalisation argument
@GitGud The part about "finish" in the question doesn't refer to the process of changing the diagonal digits but to the prior step of "listing all numbers in $\mathbb N$."
Apr
4
comment Contructive implications of Cantor's diagonalisation argument
First, Cantor's argument constructively proves that, given any sequence (indexed by $\mathbb N$) of real numbers, there is a real number not in the sequence. Second, what you call "the first part" of the proof is actually part of the hypothesis of the theorem. Third, "can one actually finish" is not a mathematical question unless "one" is a mathematical entity (in contrast to a person). Fourth, your main concern seems to be about the mere existence of $\mathbb N$, not about Cantor's proof.
Apr
4
comment If $\lim_{x\to0} f(x)+g(x)$ and $\lim_{x\to0} f(x)g(x)$ exist simultaneously, are there any $f(x)$ and $g(x)$ that do not have limit
In fact, any example that works for the original problem also works for the comment. If both $f(x)+g(x)$ and $f(x)-g(x)$ had limits as $x\to 0$, then so would their sum $2f(x)$ and their difference $2g(x)$.