Let $X$ be a separable Hausdorff space. Prove that $X ≤ 2^mathfrak{c}$
Tag: separable
functional analysis – For a compact operator $T in K(X, H)$ and H an Hilbert space, $overline{T(X)}$ is separable
Let X be a normed space, H be a Hilbert space, and let $T in K(X, H)$ the set of compact operators. Show that T(X) is separable. I tried to use the fact that the set of finite rank operators $F(X,H) = K(X,H)$ when H is an Hilbert space, to show that $T(x)$ is countable (since we need to show that $overline{T(X)}$ is separable, that is, that it contains a countable dense subset)
Universality in the class of separable Banach algebras
Let us consider the class of Banach algebras with homomorphisms that are bounded below but not necessarily isometric.

Is there are separable Banach algebra that contains isomorphic images of all separable Banach algebras?

Is there a commutative separable Banach algebra that contains commutative separable Banach algebras?
The trick with bounded the distance between commuting projections (of arbitrary norm) does not work in either case.
What does it mean for a joint distribution to be separable?
Given the joint distribution of two imagepixels at an offset v from each other modeled as
$Y(x_1, x_2; v) = Prob(f(u)=x_1 $ and $ f(u+v)=x_2)$
what does it mean to say that the joint distribution is not separable for any $v$?
This may be more of a general question of separable joint distributions, but I am guessing that it’s related to the function and whether it is separable? Can anyone provide examples of this? I’m finding it difficult to find answers elsewhere that aren’t muddled with a lot of irrelevant subject matter to weed through.
separable equation problem – Mathematics Stack Exchange
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real analysis – Is every metric space with countable base a separable metric space?
Theorem: Every compact metric space $K$ has a countable base and therefore $K$ is separable.
A: Since every infinite subset of $K$ has a limit point in $K$ as $K$ is compact, it follows that $K$ is separable and since every separable metric space has a countable base so does $K$. This proves the theorem.
Now, let’s say that this A is not known to us so we proceed as below:
Choose $delta gt 0$,
Clearly $Ksubseteq cup_{xin K} N_delta (x)$. Since, $K$ is compact, there must exist a finite subset $K_fsubset K$ such that $Ksubseteq cup_{xin K_f} N_delta (x)=B$
Let $Gsubseteq K$ be any open subset of $K$. For some $tin G$, $exists epsilon_tgt 0$ such that $N_{epsilon_{t}}(t)subset G$.
Clearly, $tin B$ and hence $d(t,x)lt delta$ for some $xin K_f$
If $ain N_delta (x)$ then $d(a,x)lt delta$
Hence, $d(a,t)le d(a,x)+d(x,t)lt 2delta$
Since $deltagt 0$ is arbitrary, we choose $delta =epsilon_t/2implies d(a,t)lt epsilon_timplies ain G$. Thus we have proved that for any $xin G, x in N_delta (x) text{(for some $delta gt 0$)}subset G$. Hence, $B$ is a countable base of $K$.
I am stuck at proving (if this is true at all):
$K$ has a countable base $implies K$ is separable. If this statement is not true, then $A$ can be employed to prove $K$ is separable.
Please help. Thanks.
galois theory – Infinite separable extensions
Let $L/K$ be an infinite algebraic extension of fields. One assumes that the fields are embedded in an algebraic closure $Omega$. Consider an element $alpha$ of $Omega$ separable over $L$. Is $K(a)$ a separable extension of $K$?
When $L/K$ is a finite extension, it is pretty easy to prove. But what about in the infinite dimensional case?
Are all first order linear differential equations separable?
All first order linear differential equations are solvable thanks to the existence of the integrating factor
But is it correct to say that all such DEs are separable? I’m not sure if the method involving the integral factors counts as solving by separation.
I ask because the “separable” adjective seems to be important in classifying DEs, and saying that all first order linear differential equations are separable is a pretty meaningful statement
functional analysis – If $ X $ is separable, $ D $ is a countable symmetric subset of the unit sphere $ B ^ * $ in $ X ^ * $ dense with respect to the Mackey topology
To let $ X $ be a separable Banach room. $ X ^ * $ will denote the dual of $ X $. $ B ^ * $ denotes the unit sphere in $ X ^ * $.
$ X ^ * $ equipped with the Mackey topology $ tau (X ^ *, X) $i.e. the strongest locally convex topology $ dau $ on $ X ^ ∗ $ for which we have $ (X ^ *, tau) ^ ∗ = X $.
Why is there a countable subset? $ D $ of the unity ball $ B ^ * $symmetrical $ (D = D) $ so that:
$$
B ^ * = overline {D} ^ { tau (X ^ *, X)}
$$
Functional analysis – search for a separable, closed subspace.
I am dealing with the following problem from Fabian's book:
To let $ X, Y $ Banach spaces and $ mathrm T $$ in $$ mathcal B $$ (X, Y) $, Show that if $ Y $ is separable and $ mathrm T $ is on $ Y $then there is a separable closed subspace $ Z $ of $ X $ so that $ mathrm T $($ Z $) =$ Y $,
I found out that I need to point out a dense, countable subset of Y, but after that I'm stuck. Any ideas?