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May
13
answered Well-defined and Equivalence relations
May
12
accepted Any finite index subgroup of $\mathbb Z_p$ is open
May
8
answered Intuitively understanding $\sum_{i=1}^ni={n+1\choose2}$
May
7
comment What would a tesseract actually look like?
Check out the hypercube intuition builder: hypersolid.milosz.ca
May
4
comment Famous Problems the Experts Could not Solve
Yitang Zhang proved his result on bounded prime gaps at the age of 58 (or thereabouts). 'Less accomplished', maybe, but certainly not 'younger'.
May
3
answered Any finite index subgroup of $\mathbb Z_p$ is open
May
3
comment Any finite index subgroup of $\mathbb Z_p$ is open
For example, what's stopping some element $x\in\mathbb Z_p$ from getting arbitrarily close to elements of $H$ - in the sense that for all $\epsilon>0$ there is some $h_\epsilon\in H$ with $|x-h_\epsilon|_p<\epsilon$, but not being itself an element of $H$? Such an element $x$ would be contained in $\bigcap H_n$ but not in $H$.
May
3
comment Any finite index subgroup of $\mathbb Z_p$ is open
Thanks for the rewrite. I'm still not sure, though, why we have that $H=\bigcap_{n\in\mathbb N} H_n$.
May
3
comment Any finite index subgroup of $\mathbb Z_p$ is open
Please could you explain the last part of this? I can see why $(a_n($ is a non-decreasing sequence and bounded by $[\mathbb Z_p:X]$, but it's not clear to me why that means that the constant value the sequence attains is equal to $[\mathbb Z_p:H]$ or why that means that $(p^n)\subset H$.
May
3
asked Any finite index subgroup of $\mathbb Z_p$ is open
May
1
comment Preserving compactness and connectedness implies continuity for functions between locally connected, locally compact spaces?
How are you doing with the proof over $\mathbb R$? I've got a nice hint for you if you want it.
May
1
revised Preserving compactness and connectedness implies continuity for functions between locally connected, locally compact spaces?
added 4 characters in body
Apr
30
answered Why is a cartesian morphism called cartesian?
Apr
30
revised Let $(X ,\mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $Cl(A) = A\cup Bd(A)$. False!
deleted 71 characters in body
Apr
30
comment Let $(X ,\mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $Cl(A) = A\cup Bd(A)$. False!
@Ennar Yes I did.
Apr
30
comment Let $(X ,\mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $Cl(A) = A\cup Bd(A)$. False!
Is he necessarily a 'he'?
Apr
30
revised Let $(X ,\mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $Cl(A) = A\cup Bd(A)$. False!
added 5 characters in body
Apr
30
answered Let $(X ,\mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $Cl(A) = A\cup Bd(A)$. False!
Apr
30
answered If $ f : S^n \to R^n $ satisfies $f(-x)=-f(x)$ for all $x \in S^n $,then there exists $ y \in S^n $ with $f(y)=0 $
Apr
30
comment If $ f : S^n \to R^n $ satisfies $f(-x)=-f(x)$ for all $x \in S^n $,then there exists $ y \in S^n $ with $f(y)=0 $
What does the Borsuk-Ulam Theorem say?