Topological embeddings of non-compact, complete metric spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:41:39Z http://mathoverflow.net/feeds/question/22013 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22013/topological-embeddings-of-non-compact-complete-metric-spaces Topological embeddings of non-compact, complete metric spaces Robin Saunders 2010-04-21T05:07:26Z 2010-05-02T02:13:01Z <p>Given a completely metrizable space, say that it has property X if it can be embedded in some metric space such that its image is not closed. For example, the real line R can be embedded, topologically, in itself as (0,1) which is not closed. A compact space such as S^1, however, clearly cannot be embedded in any metric space in this way.</p> <ol> <li><p>Is it true that a space has property X iff it can be embedded in itself in this way? My intuition says no, but I can't think of a counterexample offhand, partly because I don't know any interesting non-compact metric spaces.</p></li> <li><p>Is property X equivalent to not being compact? If every non-compact (completely metrizable) space has a metrizable compactification then this is easy, but I don't know whether that is the case, although I suspect it might be. Which leads onto the following weaker question:</p></li> <li><p>Given a space which does not have property X, can it be embedded in a space which does? Here my intuition says yes, this should be the case, but I can't think of a more general approach for constructing a suitable embedding space than compactification which, as I said, I can't prove will give a metrizable space. I don't think I know enough topology (I've only taken a couple of basic undergraduate courses).</p></li> </ol> <p>Edit: 3. above should read "given a space which has property X, can it be embedded in a space which does not?</p> http://mathoverflow.net/questions/22013/topological-embeddings-of-non-compact-complete-metric-spaces/22026#22026 Answer by Benoît Kloeckner for Topological embeddings of non-compact, complete metric spaces Benoît Kloeckner 2010-04-21T08:01:50Z 2010-04-21T08:01:50Z <p>The answer to 1. is definitely no: a regular tree of valency at least $3$ is a counter-example. An embedding into itself must have closed image due to the combinatorial constraints. However, it can easily be embedded into a plane with non-closed image.</p> http://mathoverflow.net/questions/22013/topological-embeddings-of-non-compact-complete-metric-spaces/22031#22031 Answer by Sergei Ivanov for Topological embeddings of non-compact, complete metric spaces Sergei Ivanov 2010-04-21T08:56:30Z 2010-04-24T07:13:19Z <p>The answer to 2 is yes. Let $(X,d)$ be complete but not compact, then for some $r>0$ it has a countable r-separated subset <code>$S=\{s_i\}:i=1,2,\dots$</code>. Let $d'$ be the maximal metric on $X$ satisfying $d'\le d$ and $d'(s_i,s_j)\le r/\min(i,j)$ for all $i,j$. Then $(X,d')$ is homeomorphic to $(X,d)$ - in fact, the identity map is a local isometry - but $d'$ is not complete. So we have embedded the space into the completion of $(X,d')$ as a non-closed subset.</p> <p>The metric $d'$ can be constructed explicitly: $d'(x,y)$ is the minimum of $d(x,y)$ and the infimum of sums $d(x,s_i)+d(x,s_j)+r/\min(i,j)$ over all pairs of $i,j$. Verifying the triangle inequality is straightforward.</p> <p>As for 3, the answer is no, because you cannot embed any complete space into a compact space. For example, a non-separable Banach space cannot be so embedded, as Qiaochu Yuan explained in comments.</p> <p><strong>Update.</strong> It seems that I misunderstood Q3. As stated, it asks whether every compact space can be embedded into a complete non-compact one. The answer is of course yes, as Ady noticed.</p> http://mathoverflow.net/questions/22013/topological-embeddings-of-non-compact-complete-metric-spaces/22404#22404 Answer by Ady for Topological embeddings of non-compact, complete metric spaces Ady 2010-04-24T01:33:23Z 2010-05-02T02:13:01Z <p>Any metric space can be isometrically embedded into some Banach space <em>E</em>. And <em>E</em> has the property X, since it can be topologically embedded into itself as its open unit ball, which is clearly not closed in <em>E</em>. Therefore, the answer to <em>3.</em> is <strong>YES</strong>, and the intuition of Robin Saunders is very good.</p> <hr> <p>This is just to clarify some things. For this, let $(M,d)$ be a complete metric space.</p> <p>It is pretty clear that a compact metric space cannot have "the property X". Conversely, asumming that $M$ is non-compact and that $d$ is bounded, by taking into consideration the equivalent metric $d^{*}\left(x,y\right)$= $\sup_{t\in M}$ $\mid f(x)\cdot d(x,t)$ $-f\left(y\right)\cdot d\left(y,t\right)\mid$, where $f:M$ $\rightarrow$ $(0,\infty)$ is bounded and continuous and inf$_{M}$ $f$ = 0, one easily obtain that $(M,d^*)$ is not complete.</p> <p>Consequently, $M$ has "the property X" iff it is non-compact.</p> <p>Now, every separable metric space can be embedded into the Hilbert cube, which has not "the property X" .</p> <p>[Note that on <a href="http://en.wikipedia.org/wiki/Hilbert_cube" rel="nofollow">Wikipedia</a> you'll find "every separable metric space can be isometrically embedded into the Hilbert cube", which is obviously not true.]</p> <p>OTOH, any subspace of a separable metric space is separable, too. And any compact metric space is, clearly, separable.</p> <p>Therefore, $M$ is embeddable into a space having not "the property X" iff $M$ is separable.</p>