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age | 24 | |
visits | member for | 5 years, 1 month |
seen | 1 hour ago | |
stats | profile views | 11,896 |
Nov 21 |
awarded | Enlightened |
Nov 21 |
awarded | Nice Answer |
Nov 21 |
revised |
When does $\mathbf{Top}/X$ embedd fully faithfully into $\mathbf{Top}$?
added 3 characters in body |
Nov 21 |
answered | When does $\mathbf{Top}/X$ embedd fully faithfully into $\mathbf{Top}$? |
Nov 15 |
revised |
Every free abelian group is slender, why?
deleted 55 characters in body |
Nov 15 |
answered | Every free abelian group is slender, why? |
Nov 15 |
reviewed | Leave Open Every homomorphism from the Baerâ€“Specker group into a slender group factors through ${\bf Z}^n$, why? |
Nov 15 |
reviewed | Leave Open Every free abelian group is slender, why? |
Nov 4 |
comment |
Finding a norm on $ \mathbb{R}^X $ such that the “natural” embedding of a metric space $ X $ in $ \mathbb{R}^X $ becomes an isometry
If the image of the Kuratowski embedding fails to be linearly independent, you should be able to tweak it a bit to fix this. For instance, you could enlarge $X$ to $X\times \mathbb{R}$ with (say) the $L^2$ product metric, and I think that should cause any linear dependences you may have had on $X$ to be violated. The only way I can imagine finding a suitable norm on $\mathbb{R}^X$ is by solving your original problem and then using a nonconstructive linear isomorphism between $\mathbb{R}^X$ and your Banach space--as I said before, defining a norm on all of $\mathbb{R}^X$ is hard. |
Nov 4 |
comment |
Finding a norm on $ \mathbb{R}^X $ such that the “natural” embedding of a metric space $ X $ in $ \mathbb{R}^X $ becomes an isometry
It seems to me like a far more natural choice of $T$ would be $T(x)(y)=d(x,y)$. If $X$ is bounded, this is an isometry with respect to the sup norm on the space of bounded functions from $X$ to $\mathbb{R}$. In general, if $X$ is infinite, I would not expect there to be any natural norm that is well-defined on all of $\mathbb{R}^X$ (in particular, there does not exist a norm that makes every projection continuous). |
Nov 4 |
answered | Does a graded vector space isomorphism between the homology of two loop spaces imply the existence of an algebra isomorphism? |
Nov 4 |
comment |
Order dimension vs topological dimension of a poset
Do you have any reason to think they are related? They seem wildly different. Note that a totally ordered set (such as $\omega_1$) can have uncountable covering dimension. |
Nov 3 |
answered | functions which covers(good covers) manifolds |
Nov 1 |
comment |
Self-similarity for simple algebraic structures
If you mean to ask more questions than the one question you have at the end (which is only even related to about a third of your post), you should state them explicitly. |
Oct 31 |
comment |
Relations between characteristic classes of a group and the Stiefel-Whitney/Pontryagin classes
I think you misread the question--as I read it, it is asking about relations between the characteristic classes of the tangent bundle of a manifold and characteristic classes of arbitrary $G$-bundles. |
Oct 25 |
awarded | Nice Answer |
Oct 23 |
reviewed | Leave Open Is a group uniquely determined by the sets $\{ab,ba\}$ for each pair of elements a and b? |
Oct 23 |
answered | Are there examples of families of objects which are canonically isomorphic, but where diagrams of canonical isomorphisms don't commute? |
Oct 22 |
reviewed | Leave Open A metric associated with a continuous surjective map $f:X\to Y$ |
Oct 22 |
answered | A metric associated with a continuous surjective map $f:X\to Y$ |