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1d
comment checking definition of bounded linear function involves operator maps between different spaces
@math101: Well, that's what you should do first. I will not give any more detailed answer, especially while you're not even showing any effort of your own.
1d
comment How is an empty set truly “empty”?
"In that sense, we're saying that an empty bag is a bag that contains exactly what an empty bag contains: nothing at all." -- this is a rather unfortunate expression. After all, for any set $A$ you have $\emptyset\subseteq A$, but it is not all that common for $A$ to contain nothing at all.
1d
revised checking definition of bounded linear function involves operator maps between different spaces
added 19 characters in body
1d
comment checking definition of bounded linear function involves operator maps between different spaces
Which part do you not understand? What $y^*$ is?
1d
answered checking definition of bounded linear function involves operator maps between different spaces
1d
comment checking definition of bounded linear function involves operator maps between different spaces
Still, your choice of symbols is rather sloppy here. Both $K$ and $H$ have two quite distinct meanings here. It doesn't matter much if you do keep control over it (and this being a small piece of math, it is not that hard), but be careful.
2d
comment Help show all compact sets are closed in the compact complement topology
At a glance, I think @AustinMohr made a mistake there. I believe $\tau'$-compact sets are precisely $\tau$-closed sets (sketch of proof: if a set is not $\tau$-closed, you can find an intersection with a compact set which is still not closed; this compact set will be $\tau'$-closed, and the rest should be straightforward) as well as the other way around: $\tau'$-closed sets are precisely the $\tau$-compact sets (this is trivial, though). The same should happen in any locally compact space. Then again, I may be the one who made a mistake, as I did not think much.
2d
comment Help show all compact sets are closed in the compact complement topology
Are you sure this is what you were meant to prove? A similar-sounding statement which is, I believe, true here is that every $\tau$-closed set is $\tau'$-compact. Now if you look at a $\tau$-closed set which is unbounded (and not the whole space), then it will not be $\tau'$-closed: every $\tau'$-open (nonempty) set has a bounded complement.
Aug
30
comment Only once differentiable
Why can't the derivative be discontinuous? A priori it can be discontinuous on a meagre (but very infinite, e.g. full measure) set. In which case the initial function does not have to be (again a priori) an integral, as FTC does not apply.
Aug
30
reviewed Close Find the natural number $n>2$ such that $\frac{n!}{(n-1)!} + \frac{n!}{3!(n-3)!} = 2\frac{n!}{2!(n-2)!}$
Aug
30
reviewed Leave Open Looking to attain fluency in mathematics, not academic mastery
Aug
30
reviewed Close CPT exam Quantitative aptitude exercise 8c
Aug
28
comment Is there anything wrong with this proposed proof of the irrationality of Euler's constant?
@CarstenS: The simplest way to see it is probably by noticing that the logarithm of left hand side is just $\log(2)\cdot (\sum_{n\geq 1} 2^{-n})$
Aug
26
comment Is it true that $A \in A$?
You did not define the set $A$.
Aug
23
comment How to solve an exponential and logarithmic system of equations?
This "answer" says nothing OP did not already know.
Aug
19
comment What is “algebra” in $\sigma$-algebra (or “field” in $\sigma$-field)?
@BCLC: I've missed the obvious: in algebra, an ($R$-)algebra is also often not a field. In fact, there are far fewer contexts when the two are synonymous.
Aug
19
comment What is “algebra” in $\sigma$-algebra (or “field” in $\sigma$-field)?
@BCLC: offhand, in universal algebra (there an ``algebra'' just a structure with some operations) or in functional analysis (a Banach algebra is almost never a field). And of course Boolean algebras are not in general called fields, as far as I know (except for fields of sets), much less Heyting algebras and similar objects in algebraic logic.
Jul
28
comment Is every Hilbert space an $L^2$ space?
The same how? This is more or less what this question is about...
Jul
27
comment Is every Hilbert space an $L^2$ space?
The last one is equality, not isomorphism. The congruence follows from the fact that two Hilbert spaces of the same dimension are isomorphic (which is immediate from the definition). I believe it is entirely elementary, calling it a theorem is a gross overstatement. Basically the only things this argument uses are the definitions and the fact that every Hilbert space has a (Hilbert) basis.
Jul
27
comment Is every Hilbert space an $L^2$ space?
@user51514: What theorem?