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Jun
5
comment Topological Entropy of $T$, on a disjoint union?
if, say, $A$ is not $T$-invariant, what do you mean by $h(A,T_{|A})$? You need $A$ to be $T$-invariant for $T_{|A}$ to define a self-map of $A$. So, I think the question makes sense just under the assumption that your three spaces are invariant, so that you have well-defined restrictions.
Jun
3
comment Using Exchange Lemma in an decomposition
I do not think that this is even true: take a local ring $R$, let $M=R$, so that $M^{(3)}\cong R\oplus R\oplus R$, here $B=R\oplus R$. Then, $R\oplus R$ is not a direct summand of $R$...
Jun
1
revised Quotients vs equivalence relations
added 123 characters in body
Jun
1
asked Quotients vs equivalence relations
May
13
accepted Compact subgroups are contained in open compact subgroups in locally profinite groups
May
13
comment Compact subgroups are contained in open compact subgroups in locally profinite groups
Now I see, thanks! I think your argument perfectly works now. There is a misprint in your answer: on the 5th line you claim that $C$ is open, I think you mean either $N$ or $CN$, any of the two would work.
May
12
comment Compact subgroups are contained in open compact subgroups in locally profinite groups
I do not see why the proof in your comment works, it seems a circular argument: to show that $y^{-1}Ny\subseteq N$ for a given $y\in C$, you use that $(yx)^{-1}N(yx)\subseteq N$ for all $x\in C$. But for $x=1$ this was the statement you wanted to prove...
May
12
comment Compact subgroups are contained in open compact subgroups in locally profinite groups
Thanks for posting this proof. Let me ask for some details: why is $CN$ a subgroup? Consider two elements $ck^x$ and $dh^y$, with $c,d,x,y\in C$, $h,k\in K$ and $k^x,h^y\in N$. Then, there exists $k'\in K$ such that $k^x=(k')^{yd^{-1}}$, so that $ck^x dh^y=c(k')^{yd^{-1}}dh^y=(cd)(k')^y$. Still, why is this $(k')^y$ belonging to $N$?
May
11
asked Compact subgroups are contained in open compact subgroups in locally profinite groups
Apr
29
comment $R/I^n$ is a local ring
you are right! There is something missing in my proof...
Apr
29
answered $R/I^n$ is a local ring
Apr
29
comment Choice of a skeleton
OK, thanks anyway :-)
Apr
29
comment Choice of a skeleton
Not just that. It is a full subcategory where any two isomorphic objects are equal. The part which is not clear to me is especially the second part of the question, the existence of that specific functor
Apr
29
comment Choice of a skeleton
Of course, me neither... this is why I am asking here...
Apr
29
asked Choice of a skeleton
Apr
17
answered Definition of multiplicity
Apr
4
comment Definition of multiplicity
See the limit formula of Lech for multiplicities
Jan
29
comment Verifying a Vector Space Via Given Axioms
The proof of 3) is the following: we have to show that there exists a neutral element for the sum in our candidate vector space. We affirm that $\{z_n\}$, with $z_n=0$ for all $n$, is our neutral element. We have already seen that addition is commutative so it is enough to verify that $\{\alpha_n\}+\{z_n\}=\{\alpha_n\}$ for all $\{\alpha_n\}$ (otherwise one should also check that $\{z_n\}+\{\alpha_n\}=\{\alpha_n\}$). So, $\{\alpha_n\} + \{z_n\} = \{\alpha_n + z_n\} = \{\alpha_n\}$, since $z_n=0$ is the neutral element for the addition in $\mathbb K$.
Jan
29
comment Verifying a Vector Space Via Given Axioms
The proof of 2) goes like this: $\{\alpha_n\} + \{\beta_n\}=\{\alpha_n+\beta_n\}\overset{(*)}{=}\{\beta_n+\alpha_n\} = \{\beta_n\} + \{\alpha_n\}$. Where $(*)$ holds by the commutativity of addition in $\mathbb K$.
Jan
28
comment Verifying a Vector Space Via Given Axioms
For property 7: I do not see the utility of writing $(\lambda)(\mu)$, instead of $\lambda\mu$, they mean exactly the same. What you should use here is the associativity of multiplication in the base field, do you see how? Property 8 seems OK.