Timeline for Finding a norm on $ \mathbb{R}^X $ such that the "natural" embedding of a metric space $ X $ in $ \mathbb{R}^X $ becomes an isometry
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 22, 2015 at 17:15 | vote | accept | Ormi | ||
| Nov 4, 2014 at 19:54 | answer | added | Yemon Choi | timeline score: 6 | |
| Nov 4, 2014 at 19:47 | comment | added | Ormi | I'm not asking this question with a "please solve my homework" intention, because I'm going to just keep trying to work out the linear independence for Kuratowski's embedding. I'm just genuinely curious about what can be achieved with my initial idea, so if you could give me some hints for that, or refer me to somewhere where I can read up about it, I'd be grateful. I also thought about defining the norm on the subspace of functions with finite supports, could this be the subspace you mentioned? | |
| Nov 4, 2014 at 19:17 | comment | added | Yemon Choi | Well if you were asked to do it then I am not sure I should provide a detailed answer... I suspect that the Kuratwowski embedding is the answer that was expected, but actually there is another embedding that is more or less the one you thought of, although the norm is only defined on a subspace and is more complicated to describe. | |
| Nov 4, 2014 at 19:16 | comment | added | Eric Wofsey | 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, 2014 at 19:06 | comment | added | Ormi | planetmath.org/kuratowskisembeddingtheorem This seems to be relevant here. | |
| Nov 4, 2014 at 19:03 | comment | added | Ormi | I was asked to prove that every metric space can be isometrically embedded in a Banach space, so that the image is linearly independent. My first idea was the one I described, since the linear independence is obvious there, and if a good norm could be found, any normed space be embedded in a Banach space, so that would give the desired result. Kuratowski's embedding seems to do the job much better(though I'm still not sure about the linear independence), but I found it curious to see if it was possible to find the right norm here and what the technique for doing it would be. | |
| Nov 4, 2014 at 18:47 | comment | added | Yemon Choi | The answer to your question is "almost yes" but I'm curious to know in what context this question arose. Were you asked to find such a norm, or did you read somewhere that such a norm exists? | |
| Nov 4, 2014 at 18:17 | comment | added | Ormi | Yes, d(x,y) is a much better embedding of X in in $ R^X $, but my initial idea was the embedding described above and even though it wasn't the best one, I found the question of existence of a suitable norm(no matter how "weird" it could potentially be) quite interesting in itself. | |
| Nov 4, 2014 at 17:51 | comment | added | Eric Wofsey | 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, 2014 at 17:44 | review | First posts | |||
| Nov 4, 2014 at 17:54 | |||||
| Nov 4, 2014 at 17:39 | history | asked | Ormi | CC BY-SA 3.0 |