Mixing
Continuous injections of nice spaces into $ell^2$
0
$begingroup$
If X is a separable, first countable Hausdorff space, then does there exist a continuous map from X to $ell^2$?
Intuitively put, can you always fill in the holes of nice spaces?
general-topology separable-spaces
asked Jan 19 at 9:26
AIM_BLBAIM_BLB
2,5122819
$endgroup$
add a comment |
0
$begingroup$
If X is a separable, first countable Hausdorff space, then does there exist a continuous map from X to $ell^2$?
Intuitively put, can you always fill in the holes of nice spaces?
general-topology separable-spaces
asked Jan 19 at 9:26
AIM_BLBAIM_BLB
2,5122819
$endgroup$
$begingroup$
You cannot have an embedding, but you want the map to be injective? "into" is ambiguous. You always have a constant map to $ell^2$ so you need some condition.
$endgroup$
– Henno Brandsma
Jan 19 at 9:29
$begingroup$
No, not an embedding, just a continuous injection
$endgroup$
– AIM_BLB
Jan 19 at 9:33
1
$begingroup$
If your conclusion is true, then the continuous functions on the space separate the points. I don't think that this is true of every separable first countable Hausdorff space, but I might be wrong. You should look for Urysohns theorem and the context to see conditions under which this holds.
$endgroup$
– PhoemueX
Jan 19 at 9:37
add a comment |
0
0
0
$begingroup$
If X is a separable, first countable Hausdorff space, then does there exist a continuous map from X to $ell^2$?
Intuitively put, can you always fill in the holes of nice spaces?
general-topology separable-spaces
asked Jan 19 at 9:26
AIM_BLBAIM_BLB
2,5122819
$endgroup$
If X is a separable, first countable Hausdorff space, then does there exist a continuous map from X to $ell^2$?
Intuitively put, can you always fill in the holes of nice spaces?
general-topology separable-spaces
general-topology separable-spaces
asked Jan 19 at 9:26
AIM_BLBAIM_BLB
2,5122819
asked Jan 19 at 9:26
AIM_BLBAIM_BLB
2,5122819
edited Jan 19 at 9:30
AIM_BLB
asked Jan 19 at 9:26
AIM_BLBAIM_BLB
2,5122819
asked Jan 19 at 9:26
AIM_BLBAIM_BLB
2,5122819
asked Jan 19 at 9:26
AIM_BLBAIM_BLB
2,5122819
2,5122819
$begingroup$
You cannot have an embedding, but you want the map to be injective? "into" is ambiguous. You always have a constant map to $ell^2$ so you need some condition.
$endgroup$
– Henno Brandsma
Jan 19 at 9:29
$begingroup$
No, not an embedding, just a continuous injection
$endgroup$
– AIM_BLB
Jan 19 at 9:33
1
$begingroup$
If your conclusion is true, then the continuous functions on the space separate the points. I don't think that this is true of every separable first countable Hausdorff space, but I might be wrong. You should look for Urysohns theorem and the context to see conditions under which this holds.
$endgroup$
– PhoemueX
Jan 19 at 9:37
add a comment |
$begingroup$
You cannot have an embedding, but you want the map to be injective? "into" is ambiguous. You always have a constant map to $ell^2$ so you need some condition.
$endgroup$
– Henno Brandsma
Jan 19 at 9:29
$begingroup$
No, not an embedding, just a continuous injection
$endgroup$
– AIM_BLB
Jan 19 at 9:33
1
$begingroup$
If your conclusion is true, then the continuous functions on the space separate the points. I don't think that this is true of every separable first countable Hausdorff space, but I might be wrong. You should look for Urysohns theorem and the context to see conditions under which this holds.
$endgroup$
– PhoemueX
Jan 19 at 9:37
$begingroup$
You cannot have an embedding, but you want the map to be injective? "into" is ambiguous. You always have a constant map to $ell^2$ so you need some condition.
$endgroup$
– Henno Brandsma
Jan 19 at 9:29
$begingroup$
You cannot have an embedding, but you want the map to be injective? "into" is ambiguous. You always have a constant map to $ell^2$ so you need some condition.
$endgroup$
– Henno Brandsma
Jan 19 at 9:29
$begingroup$
No, not an embedding, just a continuous injection
$endgroup$
– AIM_BLB
Jan 19 at 9:33
$begingroup$
No, not an embedding, just a continuous injection
$endgroup$
– AIM_BLB
Jan 19 at 9:33
1
1
$begingroup$
If your conclusion is true, then the continuous functions on the space separate the points. I don't think that this is true of every separable first countable Hausdorff space, but I might be wrong. You should look for Urysohns theorem and the context to see conditions under which this holds.
$endgroup$
– PhoemueX
Jan 19 at 9:37
$begingroup$
If your conclusion is true, then the continuous functions on the space separate the points. I don't think that this is true of every separable first countable Hausdorff space, but I might be wrong. You should look for Urysohns theorem and the context to see conditions under which this holds.
$endgroup$
– PhoemueX
Jan 19 at 9:37
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Let $X$ be the Double Arrow space, $[0,1] times {0,1}$ in the lexicographic order topology. Then $X$ is a non-metrisable, compact, first-countable, separable, Hausdorff (even hereditarily normal, being an ordered topological space) and if a 1-1 continuous function $f: X to Y$ existed where $Y$ is metrisable, then $f$ would be a homeomorphism (as $Y$ is Hausdorff and so $f$ is closed etc.) and hence metrisable too. So from $X$ there can be no injective map into $ell^2$.
answered Jan 19 at 11:12
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Nice counter example, I'm wondering if metrizability essential for a counterexample to be possible?
$endgroup$
– AIM_BLB
Jan 19 at 12:49
1
$begingroup$
@AIM_BLB It's the easiest: a space $X$ has an injection into $ell^2$ implies it has a coarser metrisable topology (i.e. it's submetrisable, as this is called). It is well-known that every separable metrisable space has an embedding into $ell^2$.
$endgroup$
– Henno Brandsma
Jan 19 at 12:58
$begingroup$
Oh yea? Is it possible to have a link to this paper? This would help so much
$endgroup$
– AIM_BLB
Jan 19 at 20:01
1
$begingroup$
@AIM_BLB the embedding fact is quite well-known, it’s in Bessaga and Pelczynski’s book, or in van Mill’s book on infinite-dimensional topology. It’s originally due to Banach IIRC. It uses a countable family of separating functions.
$endgroup$
– Henno Brandsma
Jan 19 at 21:42
$begingroup$
Perfect, thanks so much Henno; this makes a big difference for me :)
$endgroup$
– AIM_BLB
Jan 19 at 21:45
add a comment |
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$begingroup$
Let $X$ be the Double Arrow space, $[0,1] times {0,1}$ in the lexicographic order topology. Then $X$ is a non-metrisable, compact, first-countable, separable, Hausdorff (even hereditarily normal, being an ordered topological space) and if a 1-1 continuous function $f: X to Y$ existed where $Y$ is metrisable, then $f$ would be a homeomorphism (as $Y$ is Hausdorff and so $f$ is closed etc.) and hence metrisable too. So from $X$ there can be no injective map into $ell^2$.
answered Jan 19 at 11:12
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Nice counter example, I'm wondering if metrizability essential for a counterexample to be possible?
$endgroup$
– AIM_BLB
Jan 19 at 12:49
1
$begingroup$
@AIM_BLB It's the easiest: a space $X$ has an injection into $ell^2$ implies it has a coarser metrisable topology (i.e. it's submetrisable, as this is called). It is well-known that every separable metrisable space has an embedding into $ell^2$.
$endgroup$
– Henno Brandsma
Jan 19 at 12:58
$begingroup$
Oh yea? Is it possible to have a link to this paper? This would help so much
$endgroup$
– AIM_BLB
Jan 19 at 20:01
1
$begingroup$
@AIM_BLB the embedding fact is quite well-known, it’s in Bessaga and Pelczynski’s book, or in van Mill’s book on infinite-dimensional topology. It’s originally due to Banach IIRC. It uses a countable family of separating functions.
$endgroup$
– Henno Brandsma
Jan 19 at 21:42
$begingroup$
Perfect, thanks so much Henno; this makes a big difference for me :)
$endgroup$
– AIM_BLB
Jan 19 at 21:45
add a comment |
1
$begingroup$
Let $X$ be the Double Arrow space, $[0,1] times {0,1}$ in the lexicographic order topology. Then $X$ is a non-metrisable, compact, first-countable, separable, Hausdorff (even hereditarily normal, being an ordered topological space) and if a 1-1 continuous function $f: X to Y$ existed where $Y$ is metrisable, then $f$ would be a homeomorphism (as $Y$ is Hausdorff and so $f$ is closed etc.) and hence metrisable too. So from $X$ there can be no injective map into $ell^2$.
answered Jan 19 at 11:12
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Nice counter example, I'm wondering if metrizability essential for a counterexample to be possible?
$endgroup$
– AIM_BLB
Jan 19 at 12:49
1
$begingroup$
@AIM_BLB It's the easiest: a space $X$ has an injection into $ell^2$ implies it has a coarser metrisable topology (i.e. it's submetrisable, as this is called). It is well-known that every separable metrisable space has an embedding into $ell^2$.
$endgroup$
– Henno Brandsma
Jan 19 at 12:58
$begingroup$
Oh yea? Is it possible to have a link to this paper? This would help so much
$endgroup$
– AIM_BLB
Jan 19 at 20:01
1
$begingroup$
@AIM_BLB the embedding fact is quite well-known, it’s in Bessaga and Pelczynski’s book, or in van Mill’s book on infinite-dimensional topology. It’s originally due to Banach IIRC. It uses a countable family of separating functions.
$endgroup$
– Henno Brandsma
Jan 19 at 21:42
$begingroup$
Perfect, thanks so much Henno; this makes a big difference for me :)
$endgroup$
– AIM_BLB
Jan 19 at 21:45
add a comment |
1
1
1
$begingroup$
Let $X$ be the Double Arrow space, $[0,1] times {0,1}$ in the lexicographic order topology. Then $X$ is a non-metrisable, compact, first-countable, separable, Hausdorff (even hereditarily normal, being an ordered topological space) and if a 1-1 continuous function $f: X to Y$ existed where $Y$ is metrisable, then $f$ would be a homeomorphism (as $Y$ is Hausdorff and so $f$ is closed etc.) and hence metrisable too. So from $X$ there can be no injective map into $ell^2$.
answered Jan 19 at 11:12
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
Let $X$ be the Double Arrow space, $[0,1] times {0,1}$ in the lexicographic order topology. Then $X$ is a non-metrisable, compact, first-countable, separable, Hausdorff (even hereditarily normal, being an ordered topological space) and if a 1-1 continuous function $f: X to Y$ existed where $Y$ is metrisable, then $f$ would be a homeomorphism (as $Y$ is Hausdorff and so $f$ is closed etc.) and hence metrisable too. So from $X$ there can be no injective map into $ell^2$.
answered Jan 19 at 11:12
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 19 at 11:12
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 19 at 11:12
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 19 at 11:12
Henno BrandsmaHenno Brandsma
111k348118
111k348118
$begingroup$
Nice counter example, I'm wondering if metrizability essential for a counterexample to be possible?
$endgroup$
– AIM_BLB
Jan 19 at 12:49
1
$begingroup$
@AIM_BLB It's the easiest: a space $X$ has an injection into $ell^2$ implies it has a coarser metrisable topology (i.e. it's submetrisable, as this is called). It is well-known that every separable metrisable space has an embedding into $ell^2$.
$endgroup$
– Henno Brandsma
Jan 19 at 12:58
$begingroup$
Oh yea? Is it possible to have a link to this paper? This would help so much
$endgroup$
– AIM_BLB
Jan 19 at 20:01
1
$begingroup$
@AIM_BLB the embedding fact is quite well-known, it’s in Bessaga and Pelczynski’s book, or in van Mill’s book on infinite-dimensional topology. It’s originally due to Banach IIRC. It uses a countable family of separating functions.
$endgroup$
– Henno Brandsma
Jan 19 at 21:42
$begingroup$
Perfect, thanks so much Henno; this makes a big difference for me :)
$endgroup$
– AIM_BLB
Jan 19 at 21:45
add a comment |
$begingroup$
Nice counter example, I'm wondering if metrizability essential for a counterexample to be possible?
$endgroup$
– AIM_BLB
Jan 19 at 12:49
1
$begingroup$
@AIM_BLB It's the easiest: a space $X$ has an injection into $ell^2$ implies it has a coarser metrisable topology (i.e. it's submetrisable, as this is called). It is well-known that every separable metrisable space has an embedding into $ell^2$.
$endgroup$
– Henno Brandsma
Jan 19 at 12:58
$begingroup$
Oh yea? Is it possible to have a link to this paper? This would help so much
$endgroup$
– AIM_BLB
Jan 19 at 20:01
1
$begingroup$
@AIM_BLB the embedding fact is quite well-known, it’s in Bessaga and Pelczynski’s book, or in van Mill’s book on infinite-dimensional topology. It’s originally due to Banach IIRC. It uses a countable family of separating functions.
$endgroup$
– Henno Brandsma
Jan 19 at 21:42
$begingroup$
Perfect, thanks so much Henno; this makes a big difference for me :)
$endgroup$
– AIM_BLB
Jan 19 at 21:45
$begingroup$
Nice counter example, I'm wondering if metrizability essential for a counterexample to be possible?
$endgroup$
– AIM_BLB
Jan 19 at 12:49
$begingroup$
Nice counter example, I'm wondering if metrizability essential for a counterexample to be possible?
$endgroup$
– AIM_BLB
Jan 19 at 12:49
1
1
$begingroup$
@AIM_BLB It's the easiest: a space $X$ has an injection into $ell^2$ implies it has a coarser metrisable topology (i.e. it's submetrisable, as this is called). It is well-known that every separable metrisable space has an embedding into $ell^2$.
$endgroup$
– Henno Brandsma
Jan 19 at 12:58
$begingroup$
@AIM_BLB It's the easiest: a space $X$ has an injection into $ell^2$ implies it has a coarser metrisable topology (i.e. it's submetrisable, as this is called). It is well-known that every separable metrisable space has an embedding into $ell^2$.
$endgroup$
– Henno Brandsma
Jan 19 at 12:58
$begingroup$
Oh yea? Is it possible to have a link to this paper? This would help so much
$endgroup$
– AIM_BLB
Jan 19 at 20:01
$begingroup$
Oh yea? Is it possible to have a link to this paper? This would help so much
$endgroup$
– AIM_BLB
Jan 19 at 20:01
1
1
$begingroup$
@AIM_BLB the embedding fact is quite well-known, it’s in Bessaga and Pelczynski’s book, or in van Mill’s book on infinite-dimensional topology. It’s originally due to Banach IIRC. It uses a countable family of separating functions.
$endgroup$
– Henno Brandsma
Jan 19 at 21:42
$begingroup$
@AIM_BLB the embedding fact is quite well-known, it’s in Bessaga and Pelczynski’s book, or in van Mill’s book on infinite-dimensional topology. It’s originally due to Banach IIRC. It uses a countable family of separating functions.
$endgroup$
– Henno Brandsma
Jan 19 at 21:42
$begingroup$
Perfect, thanks so much Henno; this makes a big difference for me :)
$endgroup$
– AIM_BLB
Jan 19 at 21:45
$begingroup$
Perfect, thanks so much Henno; this makes a big difference for me :)
$endgroup$
– AIM_BLB
Jan 19 at 21:45
add a comment |
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$begingroup$
You cannot have an embedding, but you want the map to be injective? "into" is ambiguous. You always have a constant map to $ell^2$ so you need some condition.
$endgroup$
– Henno Brandsma
Jan 19 at 9:29
$begingroup$
No, not an embedding, just a continuous injection
$endgroup$
– AIM_BLB
Jan 19 at 9:33
1
$begingroup$
If your conclusion is true, then the continuous functions on the space separate the points. I don't think that this is true of every separable first countable Hausdorff space, but I might be wrong. You should look for Urysohns theorem and the context to see conditions under which this holds.
$endgroup$
– PhoemueX
Jan 19 at 9:37