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 Mar 15 awarded Enlightened Mar 15 awarded Nice Answer Mar 9 answered How to show $T$ is bijective in the following condition? Mar 9 comment Example of topological spaces $A \subset X$ with $X \setminus A$ not homeomorphic to $(X/A) \setminus (A/A)$ sorry for the ignorance, but how do you quotient a space by a subset? I mean, I can quotient by a relation (so I can quotient by particular subsets of $X\times X$), but I do not know what $X/A$ means, can you point me to a reference? Mar 9 comment Example of topological spaces $A \subset X$ with $X \setminus A$ not homeomorphic to $(X/A) \setminus (A/A)$ what is $X/A$ or $A/A$? Mar 7 comment Show if an integral domain D satisfies DCC (descending chain condition), it must satisfy ACC (ascending chain condition). well, a non-zero ideal $I$ contains a non-zero element $d$. But then $D=(d)\subseteq I\subseteq D$ implies that any non-zero ideal is $D$. (By the way, notice that a ring in which every non-zero element is a unit is a skew-field...) Mar 7 comment Show if an integral domain D satisfies DCC (descending chain condition), it must satisfy ACC (ascending chain condition). Indeed, so if non-zero element is a unit you simply don't have non-trivial ideals (so of course your Artinian domain is Noetherian). Mar 7 comment Show if an integral domain D satisfies DCC (descending chain condition), it must satisfy ACC (ascending chain condition). Hint 2: try to see $d\neq 0$ as an endomorphism of $D$ viewed as a module. The fact that $D$ is a domain tells you that this endomorphism is injective. The ideal $(d^n)$ is the image of the $n$-th power of your endomorphism... so, if $(d^n)=(d^{n+1})$ you get an isomorphism $(d^n)\to (d^{n+1})$... you should be able to conclude that multiplication by $d$ is an invertible endomorphism of $D$... so what does this tell you about the non-zero elements of $D$? Mar 6 asked On Neeman's new axiom (GTR3) for triangulated categories Feb 28 comment Arrow of Arrow Set is a Functor? I cannot see how you are defining your candidate functor. Given an arrow $f$ in ${\text{Set}}^{\to}$, you should be able to answer the following questions: Which is the image of a given set $S$? Which is the image of a given map $m:S\to S'$ between sets? If the answer to these questions is not clear, then it is complicated to understand if you are defining a functor or not... Feb 27 revised the natural map from $N_1$ to $\varinjlim N_i$ is injective? added 800 characters in body Feb 27 answered the natural map from $N_1$ to $\varinjlim N_i$ is injective? Feb 7 revised If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated. deleted 6 characters in body Feb 7 revised If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated. I simplified some parts of the reasoning following some (now disappeared) comments. Feb 7 revised If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated. Edited to make the proof independent of the choice of an order on the index set. Feb 6 comment If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated. @user26857 well, in the question they are taking a chain... which means that the modules are already ordered. Why don't you want to assume their labels are ordered? I see no restriction in that... Nov 15 awarded Yearling Sep 7 revised If $x \geq 0$, $y \geq 0$, $x \leq D$, and $0 \leq 1/(1+x) - 1/(1+y) \leq C$, then is there an upper bound on the value of $y - x$? edited the title Sep 7 suggested approved edit on If $x \geq 0$, $y \geq 0$, $x \leq D$, and $0 \leq 1/(1+x) - 1/(1+y) \leq C$, then is there an upper bound on the value of $y - x$? 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.