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Jan
16
comment An element $a$ of a monoid $M$ is invertible iff there exists $x\in M$ such that $axa=1$
Novel? It's a completely standard "trick" to write an expression and use the associative law to put parentheses two different ways to show two things are equal.
Jan
16
revised If two continuous functions are equal almost everywhere on $[a,b]$, then they are equal everywhere on $[a,b]$
added 5 characters in body
Jan
16
comment If two continuous functions are equal almost everywhere on $[a,b]$, then they are equal everywhere on $[a,b]$
TeX tip: Use \mid (instead of just |) for the set builder to get correct spacing around it.
Jan
16
comment An element $a$ of a monoid $M$ is invertible iff there exists $x\in M$ such that $axa=1$
"$x = a^{-1}a^{-1}$" has no meaning when you haven't proved that $a$ is invertible yet.
Jan
16
answered An element $a$ of a monoid $M$ is invertible iff there exists $x\in M$ such that $axa=1$
Jan
2
answered Show that the composition of an operator and its inverse is commutative.
Jan
2
comment Can any metric space be completed?
The metric on $\ell^\infty(M)$ should be $\hat d(f,g) = \lVert f-g\rVert_\infty := \sup_{x\in M} \lvert f(x) - g(x)\rvert$, right?
Jan
2
revised Can a function be continuous but not Hölder on a compact set?
deleted 3 characters in body
Jan
2
awarded  calculus
Jan
1
answered Can a function be continuous but not Hölder on a compact set?
Jan
1
comment Integral Domain but not a UFD
@SpamIAm no. $\mathbb Z[X,Y]$ is still a UFD (any polynomial ring in finitely many variables over a UFD is a UFD).
Dec
30
comment More rigorous notation than “ellipsis” for “and so forth?”
If you absolutely must use set builder notation, you should at least use proper TeX for it. Compare the output of \{a \mid P(a)\} to that of \{a|P(a)\}: $\{a \mid P(a)\}$ vs $\{a|P(a)\}$.
Dec
28
revised Learning roadmap request: compiling a “Mathematics Stack Exchange Undergraduate Bibliography”
added 188 characters in body
Dec
15
awarded  Caucus
Dec
15
comment Is $F$ Riemann integrable, then $F'$ Riemann integrable?
You are correct. There are functions which are differentiable with bounded derivative yet the derivative is not Riemann integrable. E.g. Volterra's function.
Nov
6
answered Why Open Interval In Formal Definition Of Limit At Infinity
Oct
15
comment Completion of rational numbers via Cauchy sequences
@AbstractionOfMe maybe it should be called "positivity of addition" or something. It's been so long I don't really want to edit this any more.
Oct
10
comment Closed Ball Complete iff $(M,d)$ is complete
@SujaanKunalan it seems you're not entirely clear on what the forward implication says. You're asked to prove that if $(M,d)$ is complete, then all closed balls in $M$ are complete (w.r.t. the induced metric, of course). But as I noted in my comment above, that's almost trivial. Concluding that $(M,d)$ is compact is irrelevant (and almost certainly false).
Oct
10
answered Closed Ball Complete iff $(M,d)$ is complete
Oct
10
comment Closed Ball Complete iff $(M,d)$ is complete
The forward implication can be proven much easier: a closed ball in a metric space has open complement and a closed subset of a complete space is complete.