2,812 reputation
821
bio website
location
age
visits member for 3 years, 8 months
seen 2 days ago

My profile pic is currently Sir Isaac's Newton coat of arms. It comes from wikipedia.


Oct
5
awarded  Good Answer
Oct
1
comment $\mathbb{R}P^2$ and its fundamental group by identification of edges of unity square
Only the opposite corners are identified, aren't they?
Sep
30
awarded  Explainer
Sep
30
answered Symmetric groups isomorph to dihedral groups.
Sep
29
answered The set of functions which map convergent series to convergent series
Sep
24
awarded  Autobiographer
Sep
18
comment A doubt from Milnor's “Topology from a Differentiable Viewpoint”.
But you don't want to prove that $f(M - U_1 - U_2 - \cdots - U_n)$ is open, do you? Can you see why $U - F$ is open as soon as $U$ is open and $F$ closed?
Sep
10
comment Measure of set of lines in $\mathbb{R}^2$
A union of countably many sets of measure 0 is of measure 0. So, if you have a countable set of lines, sure. If not, $A$ can be as big as you want, so no.
Sep
2
comment Error ? A subset $A$ of $ \mathbb R^p$ S.T $A^o = \phi$ and $A^- = \mathbb R^p$ where $A^o$ is interior of $A$ and $A^-$ is closure of $A$
And to prove that $\overline {\mathbb Q} = \mathbb R$, for example, take a closed set $F \subset \mathbb R$ such that $\mathbb Q \subset F$. What can you say of the complementary $\mathbb R \setminus F$?
Sep
2
comment Error ? A subset $A$ of $ \mathbb R^p$ S.T $A^o = \phi$ and $A^- = \mathbb R^p$ where $A^o$ is interior of $A$ and $A^-$ is closure of $A$
Well it's not true for all such $A$, of course. But with $A = \mathbb Q$ or $A = \mathbb R \setminus \mathbb Q$, for example, you have both $\overline A = \mathbb R$ and $A^\circ = \emptyset$.
Sep
2
comment Error ? A subset $A$ of $ \mathbb R^p$ S.T $A^o = \phi$ and $A^- = \mathbb R^p$ where $A^o$ is interior of $A$ and $A^-$ is closure of $A$
No, you will have $\overline A = \mathbb R$. Do you know the description of $\overline A$ in terms of limits of sequences?
Sep
2
comment Error ? A subset $A$ of $ \mathbb R^p$ S.T $A^o = \phi$ and $A^- = \mathbb R^p$ where $A^o$ is interior of $A$ and $A^-$ is closure of $A$
It's absolutely true that $\overline A = \mathbb R^p$ means that no proper closed subset of $\mathbb R^p$ contains $A$. But this certainly doesn't imply that $A = \mathbb R^p$. Think about $p = 1$ and $A = \mathbb Q$.
Sep
2
comment More elementary proof that $\pi_n(S^n) \cong \mathbb{Z}$
You really should read Milnor's Topology from the differentiable viewpoint: it's a jewel, it's short, and it is particulary enlightening for this isomorphism (it would be a stretch to call the proof elementary, though).
Sep
2
revised Real vector bundles on $S^{7}$
added 620 characters in body
Sep
2
comment Real vector bundles on $S^{7}$
Just some comments that you are no doubt aware of, but than can be useful to some: 1. Because $SO(n)$ is the connected component of $O(n)$ containing the identity, $\pi_k SO(n) = \pi_k O(n)$. 2. Because of Bott periodicity theorem, $\pi_6 SO(n) = 0$ forall $n \geq 8$. 3. $SO(1)$ is trivial, $SO(2)$ is a circle: no $\pi_6$ there. Conclusion: the question boils down to: "is $\pi_6 SO(n)$ nontrivial for some $n \in \{3,4,5,6,7\}$?"
Sep
2
answered Real vector bundles on $S^{7}$
Aug
28
comment Classical presentation of fundamental group of surface with boundary
You are absolutely right. Hope the edit helps.
Aug
28
revised Classical presentation of fundamental group of surface with boundary
actually answering the question
Aug
26
comment In $n>5$, topology = algebra
This is quite beautifully explained in the first chapter of Scorpan's nice The Wild World of 4-Manifolds. I recommend it strongly.
Aug
26
reviewed Approve suggested edit on Simplify the expression $(2x^2 - 2)/(x+1)$ to 2x + 2