Timeline for Are countable unions of metrizable spaces metrizable too?
Current License: CC BY-SA 3.0
18 events
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| Mar 2, 2015 at 2:16 | review | Close votes | |||
| Mar 2, 2015 at 7:27 | |||||
| Mar 1, 2015 at 17:48 | history | edited | Tomasz Kania | CC BY-SA 3.0 |
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| Jan 6, 2012 at 0:15 | vote | accept | Paul | ||
| Jan 5, 2012 at 19:25 | answer | added | Henno Brandsma | timeline score: 9 | |
| Jan 5, 2012 at 16:46 | answer | added | Stephen S | timeline score: 3 | |
| Jan 5, 2012 at 14:48 | comment | added | Pietro Majer | My excuses: the above notation for $X:=\cup_n R^n$ is completely misleading ($R^N$ is of course used for the space of all real sequences, a Frechet space, metrizable). | |
| Jan 5, 2012 at 13:44 | comment | added | Andreas Blass | For the second question, the real line is a counterexample --- metrizable, $\sigma$-compact, but not compact. (@Martin: This really is just a comment, in comparison with the other answers.) | |
| Jan 5, 2012 at 13:28 | comment | added | Pietro Majer | @Martin: sorry, you're right.. sometimes one just starts with comments this way ;-) | |
| Jan 5, 2012 at 13:24 | comment | added | Pietro Majer | @Diagonal: A sequence in the TVS $R^N$ is Cauchy (resp. convergent) if and only if, it belongs to some $R^n$ and it is Cauchy (resp., convergent) there. This implies that $R^N$ is sequentially complete. On the other hand, $R^N$ is the countable union of closed proper subspaces $R^n$ (they are closed because they are complete). Hence, $R^N$ is not metrizable by the Baire category theorem. Note that the same argument works for $C_c(\Omega)$ and $\mathcal{D}(\Omega)$ etc. | |
| Jan 5, 2012 at 13:23 | comment | added | Martin Brandenburg | Why do you all post just comments, although they are answers? | |
| Jan 5, 2012 at 13:11 | comment | added | Stephen S | Every CW-complex is the union of a sequence of closed metrizable subspaces, but there are many examples of CW-complexes that aren't metrizable (a wedge of countably infinitely many circles being perhaps the simplest). | |
| Jan 5, 2012 at 12:15 | comment | added | Buschi Sergio | Let $X=I\coprod I$ where $I=[0,1]$ the real interval by natural topolgy, then any element of $X$ is $(0, x)$ (as $x$ is in the frist copy of $I$) or $(1, x)$ (....second copy $I$). On $X$ change the topology considering the neighbors of $(0,0)$ as the usual neighbors join with a neighbor of $(1,0)$, and this family become the neighborrds of $(1,0)$ too. (formally let $f: X\to [-1, 1]: (1, x)\mapsto x,\ (0, x)\mapsto -x$ and let the topology induced by $f$ on the set $X$, where $[-1,1]$ has the natural topology). THen $X$ is a union of two subspace (metrixable of course) but $X$ isnt $T_2$. | |
| Jan 5, 2012 at 12:07 | history | edited | Paul | CC BY-SA 3.0 |
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| Jan 5, 2012 at 11:58 | comment | added | Paul | @Pietro: Why $R^N$ isn't a metrizable space? | |
| Jan 5, 2012 at 11:52 | history | edited | Paul | CC BY-SA 3.0 |
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| Jan 5, 2012 at 11:45 | comment | added | Pietro Majer | No; a classical example is $\mathbb{R}^\mathbb{N}$ as inductive limit of the finite dimensional spaces $\mathbb{R}^n$, $n\in \mathbb{N}$. Also, $C_c(\Omega)$ as inductive limit of the Banach spaces $C_K$ (=functions with support in $K$) along the family of compact subspaces $K$ of $\Omega$. | |
| Jan 5, 2012 at 11:42 | history | edited | Paul | CC BY-SA 3.0 |
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| Jan 5, 2012 at 11:35 | history | asked | Paul | CC BY-SA 3.0 |