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2h
comment What are the most prominent uses of transfinite induction outside of set theory?
This one?
3h
comment What motivates the definition of a ring in abstract algebra?
If $X$ is an Abelian group (aka a $\mathbb{Z}$-module), then $\mathrm{End}(X)$ is a ring (aka a $\mathbb{Z}$-algebra.) More generally, given a commutative ring $R$, we have that if $X$ is an $R$-module, then $\mathrm{End}(X)$ is an $R$-algebra. For example, $\mathbb{R}$-algebras of real-valued square matrices can be understood as arising in this fashion.
7h
awarded  Good Question
7h
revised (A question regarding:) the graph associated with an open cover of a topological space.
added 55 characters in body
7h
comment (A question regarding:) the graph associated with an open cover of a topological space.
@NielsDiepeveen, good point! This makes me wonder whether certain separation axioms may be needed.
7h
comment (A question regarding:) the graph associated with an open cover of a topological space.
@Stefan, no, since I'm not sure what that would mean. Do you have an idea for a definition?
7h
comment What journals/websites do mathematicians have to check frequently?
I highly doubt there is a "must-read journal" that caters to the interests of every mathematician. The question would be improved by listing the areas in which you're interested.
7h
asked (A question regarding:) the graph associated with an open cover of a topological space.
7h
accepted Does $\mathcal{P}(\alpha)$ have maximal well-founded subsets?
2d
comment Is it true that if $\kappa < \kappa^\nu$, then ${\mathrm{cf}(\kappa)} \leq \nu$?
Thanks :) $\;\!$
2d
comment Is it true that if $\kappa < \kappa^\nu$, then ${\mathrm{cf}(\kappa)} \leq \nu$?
(Wish I knew how to do smaller linebreaks.)
2d
comment Is it true that if $\kappa < \kappa^\nu$, then ${\mathrm{cf}(\kappa)} \leq \nu$?
By the previous discussion, the following is a weakening of GCH: $\tag*{}$$\tag*{}$ Axiom. (For all infinite cardinals $\kappa$ and $\nu$.) If $\kappa^\nu > \kappa$, then $\nu \geq \mathrm{cf}(\kappa)$. $\tag*{}$$\tag*{}$ I just noticed something cool: in fact, its equivalent to GCH. In particular, assume the above axiom, and suppose for a contradiction that $2^{\kappa}>\kappa^+.$ Then $(\kappa^+)^\kappa > \kappa^+$. So by the aforementioned axiom, $\kappa \geq \mathrm{cf}(\kappa^+)$, a contradiction.
2d
comment When trying to learn analysis from bottom up, what numbers should I first construct?
Thanks, you rock. Plus, now I actually know what you mean :)
2d
comment Is the cardinality of $\mathbb{Z^R}$=$\mathbb{R^Z}$?
@Secret, also, the claim that $|\mathbb{R}| < |\mathbb{R}^\mathbb{Z}|$ is incorrect, since from my answer, we have: $|\mathbb{R}| = \beth_1$ and $|\mathbb{R}^\mathbb{Z}| = \beth_1$.
2d
comment Is the cardinality of $\mathbb{Z^R}$=$\mathbb{R^Z}$?
@Secret, by the way, the "sandwich theorem" you're looking for is called the Schroder-Bernstein theorem.
2d
comment Is the cardinality of $\mathbb{Z^R}$=$\mathbb{R^Z}$?
@Secret, pretty much, but you've got the inequalities backward :)
2d
revised Is the cardinality of $\mathbb{Z^R}$=$\mathbb{R^Z}$?
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2d
comment When trying to learn analysis from bottom up, what numbers should I first construct?
I fully agree with this advice (and semirings should really be given a lot more attention.) However, I much prefer the notations $\mathbb{R}_{>0}$ and $\mathbb{R}_{\geq 0}$ to the much more ambiguous $\mathbb{R}_+$. When you write $\mathbb{R}_+$, I do not know what you mean.
2d
revised Is the cardinality of $\mathbb{Z^R}$=$\mathbb{R^Z}$?
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2d
answered Is the cardinality of $\mathbb{Z^R}$=$\mathbb{R^Z}$?