bio | website | maths.lancs.ac.uk/~kania |
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location | Lancaster, United Kingdom | |
age | ||
visits | member for | 4 years |
seen | 8 hours ago | |
stats | profile views | 1,854 |
May 15 |
awarded | Yearling |
May 8 |
awarded | Popular Question |
May 5 |
revised |
B(H) as a direct sum of a closed two sided ideal and a subalgebra
added 285 characters in body |
May 5 |
comment |
constant rank theorem for banach spaces
Yes, I meant Schauder basis too. |
May 5 |
comment |
constant rank theorem for banach spaces
@Jochen Wengenroth, a separable Banach space need not have a basis so strictly speaking these two things are not equivalent :-) |
May 1 |
revised |
Predual of a Direct Sum of Banach Spaces
added 131 characters in body |
May 1 |
comment |
Banach-Stone Theorem in Lipschitz-free spaces
I might be missing something but non-linearity doesn't seem to be essential here. Indeed, by the Mazur-Ulam theorem such map $T$ is affine and hence $T-T(0)I$ is a linear isometry. |
Apr 21 |
comment |
Characterizations of Wiener algebra
As for (3), $C^0_{(0)}$ contains a copy of $c_0$ (take the linear span of a sequence disjointly supported functions), whereas for $L_1(\mu)$ this is clearly impossible for many reasons (e.g. because $L_1(\mu)$ is weakly sequentially complete). |
Apr 17 |
reviewed | Approve What is a simplicial commutative ring from the point of view of homotopy theory? |
Mar 18 |
comment |
Images of $\{0,1\}^\kappa$
(To a now-deleted response) that cannot be right. Every subspace of $\{0,1\}^\kappa$ is zero-dimensional. As Bill says, every compact Hausdorff space is a continuous image of a zero-dimensional space and each zero-dimensional space $X$ embeds into $\{0,1\}^\kappa$ (for $\kappa$ equal to the weight of $X$). See Engelking's General topology. |
Mar 18 |
revised |
Images of $\{0,1\}^\kappa$
added 44 characters in body |
Mar 18 |
revised |
Images of $\{0,1\}^\kappa$
edited body |
Mar 18 |
answered | Images of $\{0,1\}^\kappa$ |
Mar 4 |
revised |
Communal problem books
a somewhat peculiar spelling corrected |
Mar 1 |
revised |
Are countable unions of metrizable spaces metrizable too?
added 22 characters in body; edited title |
Mar 1 |
comment |
The Banach space of bounded functions with countable support
arxiv.org/abs/1502.03026 |
Mar 1 |
comment |
What if Current Foundations of Mathematics are Inconsistent?
People were saying the same about the USSR. |
Feb 26 |
revised |
Comparing cardinalities of the spectrum of two masas in $B(H)$
will add that later |
Feb 26 |
revised |
Comparing cardinalities of the spectrum of two masas in $B(H)$
added 597 characters in body |
Feb 26 |
comment |
Comparing cardinalities of the spectrum of two masas in $B(H)$
No, if $K$ satisfies the countable chain condition (and certainly the spectrum $K$ of $L_\infty(\mu)$ for a probability measure $\mu$ does), then no subspace of $C(K)$ is isomorphic to $c_0(\omega_1)$. (This is easy modulo certain prehistoric facts from Banach space theory.) I suggest deleting the old comments. |