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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?
Jul
15
comment Is there an embedding of $\langle \mathbb{N} \setminus \{0\}, \leq, \times, 1 \rangle$ into $\langle \mathbb{N}, \leq, +, 0 \rangle$?
The reverse embedding does not exist because $0\cdot x=0$ for any $x$, while for any $x$ we have $x+1\neq x$, so there is no value for $0$.
Jun
27
comment Orthonormal Hamel Basis is equivalent to finite dimension
@Sushil: Not really. Jech's set theory probably contains everything you need to know and more, but it might not be the right text for you.
Jun
26
comment Orthonormal Hamel Basis is equivalent to finite dimension
@Sushil: No, this is not true. GCH implies that every successor cardinal is of this form, but there are other cardinals as well. This discussion is very much beyond the scope of this question. I suggest you read a text to acquaint yourself with basic cardinal arithmetic.
Jun
26
comment Orthonormal Hamel Basis is equivalent to finite dimension
@Sushil: No, this is not true for arbitrary infinite $\kappa$.
Jun
26
comment Is :$\frac{\Bbb d}{\Bbb d x}$ a chaotic operator in infinite-dimensional Hilbert space?
The derivative of a linear function is the function itself (at every point). You really need to clarify what space you have in mind and what definitions you are using.
Jun
26
comment On some equivalences of $\omega$-categoricity.
@user248653: A saturated model cannot omit a type. An infinite compact space has a non-isolated point. Alternatively, any two saturated models of the same cardinality are isomorphic.
Jun
25
comment On some equivalences of $\omega$-categoricity.
@user248653: As I have said, omitting types is crucial here.
Jun
25
comment Orthonormal Hamel Basis is equivalent to finite dimension
@Sushil: More or less, yes.
Jun
24
answered On some equivalences of $\omega$-categoricity.
Jun
21
comment Orthonormal Hamel Basis is equivalent to finite dimension
@Sushil: Hamel dimension coincides with cardinality of the space if the dimension is not smaller than the cardinality of the underlying field (so $\mathfrak c$ in case of real/complex Hilbert spaces). To see this, just recall that an infinite set is equinumerous with the set of its finite subsets. I did not say that $\kappa^\omega>\kappa$ for $\kappa>\mathfrak c$, but there are such numbers $\kappa$, including (but possibly not limited to) those of cofinality $\omega$, due to König's lemma. And example of such a number is $\beth_\omega$.
Jun
21
comment Continuous version of the Cantor-Schroder-Bernstein Theorem
That is not a duplicate per se, but actually it has a stronger assumption, and the answer is still no.
Jun
21
comment False proof why $C^1$ implies locally Lipschitz
@astudent: It's not really defined differently, it's just that the general formula does not simplify as much as it does in one dimension -- you need to specify the direction, which is not necessary in one dimension, where we have a "default" direction. And again, that is merely a superficial flaw.
Jun
21
revised False proof why $C^1$ implies locally Lipschitz
added 483 characters in body
Jun
21
answered False proof why $C^1$ implies locally Lipschitz
Jun
21
comment Can recurrences involving $\gcd$ be solved?
Anyway, if any $a_i$ is zero, and there are any solutions, there are infinitely many of them.
Jun
21
comment Can recurrences involving $\gcd$ be solved?
@HenningMakholm: This is why I asked for non-trival examples. If almost all $a_i$ are zero, then the right hand side is eventually constant, and thus must be equal to one (because there are infinitely many primes).
Jun
21
comment Can recurrences involving $\gcd$ be solved?
Anyway, it looks like it would be very hard to find $a_i$ for which the solution even exists. Do you have any non-trivial example?