Reputation
20,284
Next tag badge:
352/400 score
82/80 answers
Badges
2 40 84
Impact
~200k people reached

May
20
comment $\bigvee _{n\in\mathbb{N}}\mathbb{S}^1$ and its homeomorphisms
@Dejan: Thanks ;-)
May
20
comment $\bigvee _{n\in\mathbb{N}}\mathbb{S}^1$ and its homeomorphisms
Are you sure this is true? What if $X$ is the union of $S_n$ where $S_n$ is the circle given by the equation $(x-n)^2 + y^2 = n^2$? (Think of starting with a small circle, and drawing larger and larger circles around it, all meeting at some predetermined point.)
May
20
comment Give an example of a sequence in $l^{\infty}$ which cannot be generated by the linear operator.
I'm not sure what you mean by the supremum of a series. Yes, the sequence which consists of alternating 1s and 0s should work fine.
May
20
comment Give an example of a sequence in $l^{\infty}$ which cannot be generated by the linear operator.
The (partial sums of the ) harmonic series are unbounded.
May
20
comment Give an example of a sequence in $l^{\infty}$ which cannot be generated by the linear operator.
Your logic is getting to the right idea: You want a bounded sequence without a limit.
May
19
comment Showing two spaces are homotopy equivalent
I would define the deformation retract in pieces. First, collapse $D_2$ down to $S^2\ cup I$, then collapse $I$.
May
19
comment If the top Stiefel-Whitney class of a compact manifold is nonzer0, must there be another non-vanishing Stiefel-Whitney class?
Looks good to me. I hate it when I miss using an argument I already know. Nice catch! For some reason, I was thinking that the "$v_{n/2} = 0$ iff intersection form is even" was a $4$-manifold result.
May
19
accepted If the top Stiefel-Whitney class of a compact manifold is nonzer0, must there be another non-vanishing Stiefel-Whitney class?
May
19
asked If the top Stiefel-Whitney class of a compact manifold is nonzer0, must there be another non-vanishing Stiefel-Whitney class?
May
19
comment If $f:X \to [0,1]$ be an onto continuous map and $\{f^{-1} (y)\}$ is compact then Is $X$ compact?
No worries - I don't need the rep.
May
19
comment What is the difference between homotopy and homeomorphism?
I just received a downvote on this question (2 years after posting it). If anyone would like to suggest an improvement, I'd love to hear it....
May
19
comment If $f:X \to [0,1]$ be an onto continuous map and $\{f^{-1} (y)\}$ is compact then Is $X$ compact?
Nice answer ;-). I don't really need/want any more reputation, so I'll delete mine in a second.
May
12
comment Suppose that $A$ and $B$ are sets and that $f: A \rightarrow B$ is onto. Does being onto guarantee the sets are finite?
"There is no guarantee that an onto function sends one $x$ value to one and only one $y$ value" - actually, the "function" part of "onto function" guarantees this. Why don't you write down some examples of $A$s and $B$s and refine your conjecture?
May
8
comment Finding the fallacy in this wrong limit computing result
@g.kov: In my proof (which starts with "Now, look at the definition of ..."), I no where used any property of $0$.
May
8
comment Finding the fallacy in this wrong limit computing result
@g.kov: I think you don't understand the formal definition of limit. In a sequence, there is no last term. The sequence is literally just $-1$ repeated, and the limit is $-1$. I've given you a proof above. Unless you think there is a mistake in the proof, I have nothing else to say on the matter. You may consider asking a question on MSE to let other people weigh in, if it makes you feel better.
May
5
comment Subspace of Lindelöf space is not Lindelöf: Example
If you want a simpler example, let $X$ be any set of uncountable cardinality, and let $Y = X \cup \{x\}$ for some $x\notin X$. Declare a set to be open iff it's $Y$, empty, or a subset of $X$. Then $Y$ is compact (the only open set which contains $x$ is $Y$ itself), but the subspace $X$ is not Lindelof.
Apr
20
comment Finding the fallacy in this wrong limit computing result
Now, look at the definition of $\lim_{k\rightarrow\infty} x_k = L$. It, by definition, means that for all $\epsilon > 0$, there is an $N >0$ with the property that if $k > N$, then $|x_k - L| < \epsilon$. Now, given the constant sequence $x_k = -1$, and given any $\epsilon > 0$, take $L = -1$ and $N = 1$. Then $|x_k - L| = |-1 -(-1)| = 0 < \epsilon$. I agree that infinity is tricky business in general, but when applied to constant sequences it's not.
Apr
20
comment Finding the fallacy in this wrong limit computing result
Note that $\rho(x)$ is a slight modification of the classical Dirichlet fuction, and is well known to be discontinuous everywhere. In particular, $\lim_{k\rightarrow \infty} \rho(x_k) \neq \rho(\lim_{k\rightarrow\infty} x_k)$ in general. (What your comment effectively demonstrates is that $\rho$ is discontinuous at $x = e$).
Apr
20
comment Finding the fallacy in this wrong limit computing result
I disagree with your first equality (and now see that I missed the fact that it's $\frac{1}{2k+1}$, not $2k+1$). The point is the any (real) odd root of $(-1)$ is $-1$, so the sequence you're looking at is literally $-1,-1, -1, ...$. This has limit $-1$.
Apr
20
comment Finding the fallacy in this wrong limit computing result
Maybe I'm misunderstanding. Are you claiming that $\lim_{k\rightarrow \infty} (-1)^{2k+1} = -1$ or not? (If we are thinking of the sequential notion of a limit, that is, $k\in \mathbb{N}$, then the limit is $1$. If it's the analysis notion of a limit ($k\in \mathbb{R}$), then I agree there is no limit.)