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How to prove that compact subspaces of the Sorgenfrey line are countable?
$begingroup$
How to prove that every compact subspace of the Sorgenfrey line is countable?
general-topology compactness
edited Feb 10 '16 at 9:30
Tomek Kania
12.3k11945
asked Jul 12 '12 at 11:48
PaulPaul
14k42465
$endgroup$
add a comment |
$begingroup$
How to prove that every compact subspace of the Sorgenfrey line is countable?
general-topology compactness
edited Feb 10 '16 at 9:30
Tomek Kania
12.3k11945
asked Jul 12 '12 at 11:48
PaulPaul
14k42465
$endgroup$
add a comment |
$begingroup$
How to prove that every compact subspace of the Sorgenfrey line is countable?
general-topology compactness
edited Feb 10 '16 at 9:30
Tomek Kania
12.3k11945
asked Jul 12 '12 at 11:48
PaulPaul
14k42465
$endgroup$
How to prove that every compact subspace of the Sorgenfrey line is countable?
general-topology compactness
general-topology compactness
edited Feb 10 '16 at 9:30
Tomek Kania
12.3k11945
asked Jul 12 '12 at 11:48
PaulPaul
14k42465
edited Feb 10 '16 at 9:30
Tomek Kania
12.3k11945
asked Jul 12 '12 at 11:48
PaulPaul
14k42465
edited Feb 10 '16 at 9:30
Tomek Kania
12.3k11945
edited Feb 10 '16 at 9:30
Tomek Kania
12.3k11945
edited Feb 10 '16 at 9:30
Tomek Kania
12.3k11945
12.3k11945
asked Jul 12 '12 at 11:48
PaulPaul
14k42465
asked Jul 12 '12 at 11:48
PaulPaul
14k42465
asked Jul 12 '12 at 11:48
PaulPaul
14k42465
14k42465
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $C$ be a compact subset of the Sorgenfrey line (so $X = mathbb{R}$ with a base of open
sets of the form $[a,b)$, for $a < b$). The usual (order) topology on $mathbb{R}$ is coarser (as all open intervals $(a,b)$ can be written as unions of Sorgenfrey-open sets $[a+frac{1}{n}, b)$ for large enough $n$, so are Sorgenfrey-open as well) so $C$ is compact in the usual topology as well. This means in particular that $C$ is closed and bounded in the usual topology on $mathbb{R}$.
Suppose that $x_0 < x_1 < x_2 < ldots $ is a strictly increasing sequence in $C$, and let $c = sup {x_n: n =0,1,ldots }$, which exists and lies in $C$ by the above remarks. Also let $m = min(C)$, which also exists by the same.
Then the sets $[c,rightarrow)$ and $[m, x_0)$ (if non-empty), $[x_n, x_{n+1})$, for $n ge 0$ form a disjoint countable cover of $C$, so we cannot omit a single member of it (we need $[x_n, x_{n+1})$ to cover $x_n$, e.g.), so there is no finite subcover of it that still covers $C$.
This contradicts that $C$ is compact.
We conclude that $C$ has no infinite strictly increasing sequences. Or otherwise put: $C$ in the reverse order (from the standard one) is well-ordered.
And so we have shown that every compact subset of $C$ corresponds to a well-ordered subset of $mathbb{R}$ (by reversing the order, and note that the reals are order isomorphic to its reverse order). And all well-ordered subsets of $mathbb{R}$ are (at most) countable (this follows from several arguments, including one using second countability, e.g.).
answered Jul 14 '12 at 12:12
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
$begingroup$
why $C$ compact having no infinite strictly increasing sequences is the same as $C$ in the reverse order being well-ordered?
$endgroup$
– creepyrodent
Nov 6 '18 at 14:08
1
$begingroup$
@dude3221 A strict linearly ordered set $(X,<)$ is well-ordered (every non-empty subset has a minimum) iff there are no infinite decreasing sequences under $<$. This is classical.
$endgroup$
– Henno Brandsma
Nov 6 '18 at 14:11
add a comment |
$begingroup$
Hint: any uncountable set of real numbers contains a strictly increasing infinite sequence.
Hint 2 (added later): show that if a subspace $X$ of the Sorgenfrey line contains a strictly increasing infinite sequence, then $X$ has an open cover with no finite subcover.
answered Jul 12 '12 at 11:58
Colin McQuillanColin McQuillan
4,2681114
$endgroup$
1
$begingroup$
@John But Collin's right.
$endgroup$
– martini
Jul 12 '12 at 12:59
1
$begingroup$
You should try it yourself, for starters. And in the classical topology of $mathbb R$ such sequences can be convergent in contrast to the Sorgenfrey case, where such sequences can't have a convergent subsequence ...
$endgroup$
– martini
Jul 12 '12 at 13:06
$begingroup$
@John: see my second hint above.
$endgroup$
– Colin McQuillan
Jul 12 '12 at 13:44
$begingroup$
The Sorgenfrey line is totally disconnected. One can prove that an infinite totally disconnected space doesn't have uncountable compact subsets, even when it's not discret.
$endgroup$
– Temitope.A
Jul 20 '12 at 13:08
$begingroup$
@Temitope.A Nonsense. The Cantor set is uncountable, totally disconnected and compact.
$endgroup$
– Henno Brandsma
Jan 28 at 9:14
add a comment |
$begingroup$
not true. The compact subsets of Sorgenfrey line must be finite sets. If it is infinite countable set even it can not be compact in the ususal topology on R
answered Feb 10 '16 at 9:23
cenap ozelcenap ozel
1
$endgroup$
$begingroup$
this doesn't make any sense. [0,1] is compact in $mathbb R$ with the usual topology and is even uncountable infinite. And what has this to do with the Sorgenfrey line anyway??
$endgroup$
– noctusraid
Feb 10 '16 at 9:27
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $C$ be a compact subset of the Sorgenfrey line (so $X = mathbb{R}$ with a base of open
sets of the form $[a,b)$, for $a < b$). The usual (order) topology on $mathbb{R}$ is coarser (as all open intervals $(a,b)$ can be written as unions of Sorgenfrey-open sets $[a+frac{1}{n}, b)$ for large enough $n$, so are Sorgenfrey-open as well) so $C$ is compact in the usual topology as well. This means in particular that $C$ is closed and bounded in the usual topology on $mathbb{R}$.
Suppose that $x_0 < x_1 < x_2 < ldots $ is a strictly increasing sequence in $C$, and let $c = sup {x_n: n =0,1,ldots }$, which exists and lies in $C$ by the above remarks. Also let $m = min(C)$, which also exists by the same.
Then the sets $[c,rightarrow)$ and $[m, x_0)$ (if non-empty), $[x_n, x_{n+1})$, for $n ge 0$ form a disjoint countable cover of $C$, so we cannot omit a single member of it (we need $[x_n, x_{n+1})$ to cover $x_n$, e.g.), so there is no finite subcover of it that still covers $C$.
This contradicts that $C$ is compact.
We conclude that $C$ has no infinite strictly increasing sequences. Or otherwise put: $C$ in the reverse order (from the standard one) is well-ordered.
And so we have shown that every compact subset of $C$ corresponds to a well-ordered subset of $mathbb{R}$ (by reversing the order, and note that the reals are order isomorphic to its reverse order). And all well-ordered subsets of $mathbb{R}$ are (at most) countable (this follows from several arguments, including one using second countability, e.g.).
answered Jul 14 '12 at 12:12
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
$begingroup$
why $C$ compact having no infinite strictly increasing sequences is the same as $C$ in the reverse order being well-ordered?
$endgroup$
– creepyrodent
Nov 6 '18 at 14:08
1
$begingroup$
@dude3221 A strict linearly ordered set $(X,<)$ is well-ordered (every non-empty subset has a minimum) iff there are no infinite decreasing sequences under $<$. This is classical.
$endgroup$
– Henno Brandsma
Nov 6 '18 at 14:11
add a comment |
$begingroup$
Let $C$ be a compact subset of the Sorgenfrey line (so $X = mathbb{R}$ with a base of open
sets of the form $[a,b)$, for $a < b$). The usual (order) topology on $mathbb{R}$ is coarser (as all open intervals $(a,b)$ can be written as unions of Sorgenfrey-open sets $[a+frac{1}{n}, b)$ for large enough $n$, so are Sorgenfrey-open as well) so $C$ is compact in the usual topology as well. This means in particular that $C$ is closed and bounded in the usual topology on $mathbb{R}$.
Suppose that $x_0 < x_1 < x_2 < ldots $ is a strictly increasing sequence in $C$, and let $c = sup {x_n: n =0,1,ldots }$, which exists and lies in $C$ by the above remarks. Also let $m = min(C)$, which also exists by the same.
Then the sets $[c,rightarrow)$ and $[m, x_0)$ (if non-empty), $[x_n, x_{n+1})$, for $n ge 0$ form a disjoint countable cover of $C$, so we cannot omit a single member of it (we need $[x_n, x_{n+1})$ to cover $x_n$, e.g.), so there is no finite subcover of it that still covers $C$.
This contradicts that $C$ is compact.
We conclude that $C$ has no infinite strictly increasing sequences. Or otherwise put: $C$ in the reverse order (from the standard one) is well-ordered.
And so we have shown that every compact subset of $C$ corresponds to a well-ordered subset of $mathbb{R}$ (by reversing the order, and note that the reals are order isomorphic to its reverse order). And all well-ordered subsets of $mathbb{R}$ are (at most) countable (this follows from several arguments, including one using second countability, e.g.).
answered Jul 14 '12 at 12:12
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
$begingroup$
why $C$ compact having no infinite strictly increasing sequences is the same as $C$ in the reverse order being well-ordered?
$endgroup$
– creepyrodent
Nov 6 '18 at 14:08
1
$begingroup$
@dude3221 A strict linearly ordered set $(X,<)$ is well-ordered (every non-empty subset has a minimum) iff there are no infinite decreasing sequences under $<$. This is classical.
$endgroup$
– Henno Brandsma
Nov 6 '18 at 14:11
add a comment |
$begingroup$
Let $C$ be a compact subset of the Sorgenfrey line (so $X = mathbb{R}$ with a base of open
sets of the form $[a,b)$, for $a < b$). The usual (order) topology on $mathbb{R}$ is coarser (as all open intervals $(a,b)$ can be written as unions of Sorgenfrey-open sets $[a+frac{1}{n}, b)$ for large enough $n$, so are Sorgenfrey-open as well) so $C$ is compact in the usual topology as well. This means in particular that $C$ is closed and bounded in the usual topology on $mathbb{R}$.
Suppose that $x_0 < x_1 < x_2 < ldots $ is a strictly increasing sequence in $C$, and let $c = sup {x_n: n =0,1,ldots }$, which exists and lies in $C$ by the above remarks. Also let $m = min(C)$, which also exists by the same.
Then the sets $[c,rightarrow)$ and $[m, x_0)$ (if non-empty), $[x_n, x_{n+1})$, for $n ge 0$ form a disjoint countable cover of $C$, so we cannot omit a single member of it (we need $[x_n, x_{n+1})$ to cover $x_n$, e.g.), so there is no finite subcover of it that still covers $C$.
This contradicts that $C$ is compact.
We conclude that $C$ has no infinite strictly increasing sequences. Or otherwise put: $C$ in the reverse order (from the standard one) is well-ordered.
And so we have shown that every compact subset of $C$ corresponds to a well-ordered subset of $mathbb{R}$ (by reversing the order, and note that the reals are order isomorphic to its reverse order). And all well-ordered subsets of $mathbb{R}$ are (at most) countable (this follows from several arguments, including one using second countability, e.g.).
answered Jul 14 '12 at 12:12
Henno BrandsmaHenno Brandsma
114k348123
$endgroup$
Let $C$ be a compact subset of the Sorgenfrey line (so $X = mathbb{R}$ with a base of open
sets of the form $[a,b)$, for $a < b$). The usual (order) topology on $mathbb{R}$ is coarser (as all open intervals $(a,b)$ can be written as unions of Sorgenfrey-open sets $[a+frac{1}{n}, b)$ for large enough $n$, so are Sorgenfrey-open as well) so $C$ is compact in the usual topology as well. This means in particular that $C$ is closed and bounded in the usual topology on $mathbb{R}$.
Suppose that $x_0 < x_1 < x_2 < ldots $ is a strictly increasing sequence in $C$, and let $c = sup {x_n: n =0,1,ldots }$, which exists and lies in $C$ by the above remarks. Also let $m = min(C)$, which also exists by the same.
Then the sets $[c,rightarrow)$ and $[m, x_0)$ (if non-empty), $[x_n, x_{n+1})$, for $n ge 0$ form a disjoint countable cover of $C$, so we cannot omit a single member of it (we need $[x_n, x_{n+1})$ to cover $x_n$, e.g.), so there is no finite subcover of it that still covers $C$.
This contradicts that $C$ is compact.
We conclude that $C$ has no infinite strictly increasing sequences. Or otherwise put: $C$ in the reverse order (from the standard one) is well-ordered.
And so we have shown that every compact subset of $C$ corresponds to a well-ordered subset of $mathbb{R}$ (by reversing the order, and note that the reals are order isomorphic to its reverse order). And all well-ordered subsets of $mathbb{R}$ are (at most) countable (this follows from several arguments, including one using second countability, e.g.).
answered Jul 14 '12 at 12:12
Henno BrandsmaHenno Brandsma
114k348123
edited Jan 28 at 9:11
answered Jul 14 '12 at 12:12
Henno BrandsmaHenno Brandsma
114k348123
answered Jul 14 '12 at 12:12
Henno BrandsmaHenno Brandsma
114k348123
answered Jul 14 '12 at 12:12
Henno BrandsmaHenno Brandsma
114k348123
114k348123
$begingroup$
why $C$ compact having no infinite strictly increasing sequences is the same as $C$ in the reverse order being well-ordered?
$endgroup$
– creepyrodent
Nov 6 '18 at 14:08
1
$begingroup$
@dude3221 A strict linearly ordered set $(X,<)$ is well-ordered (every non-empty subset has a minimum) iff there are no infinite decreasing sequences under $<$. This is classical.
$endgroup$
– Henno Brandsma
Nov 6 '18 at 14:11
add a comment |
$begingroup$
why $C$ compact having no infinite strictly increasing sequences is the same as $C$ in the reverse order being well-ordered?
$endgroup$
– creepyrodent
Nov 6 '18 at 14:08
1
$begingroup$
@dude3221 A strict linearly ordered set $(X,<)$ is well-ordered (every non-empty subset has a minimum) iff there are no infinite decreasing sequences under $<$. This is classical.
$endgroup$
– Henno Brandsma
Nov 6 '18 at 14:11
$begingroup$
why $C$ compact having no infinite strictly increasing sequences is the same as $C$ in the reverse order being well-ordered?
$endgroup$
– creepyrodent
Nov 6 '18 at 14:08
$begingroup$
why $C$ compact having no infinite strictly increasing sequences is the same as $C$ in the reverse order being well-ordered?
$endgroup$
– creepyrodent
Nov 6 '18 at 14:08
1
1
$begingroup$
@dude3221 A strict linearly ordered set $(X,<)$ is well-ordered (every non-empty subset has a minimum) iff there are no infinite decreasing sequences under $<$. This is classical.
$endgroup$
– Henno Brandsma
Nov 6 '18 at 14:11
$begingroup$
@dude3221 A strict linearly ordered set $(X,<)$ is well-ordered (every non-empty subset has a minimum) iff there are no infinite decreasing sequences under $<$. This is classical.
$endgroup$
– Henno Brandsma
Nov 6 '18 at 14:11
add a comment |
$begingroup$
Hint: any uncountable set of real numbers contains a strictly increasing infinite sequence.
Hint 2 (added later): show that if a subspace $X$ of the Sorgenfrey line contains a strictly increasing infinite sequence, then $X$ has an open cover with no finite subcover.
answered Jul 12 '12 at 11:58
Colin McQuillanColin McQuillan
4,2681114
$endgroup$
1
$begingroup$
@John But Collin's right.
$endgroup$
– martini
Jul 12 '12 at 12:59
1
$begingroup$
You should try it yourself, for starters. And in the classical topology of $mathbb R$ such sequences can be convergent in contrast to the Sorgenfrey case, where such sequences can't have a convergent subsequence ...
$endgroup$
– martini
Jul 12 '12 at 13:06
$begingroup$
@John: see my second hint above.
$endgroup$
– Colin McQuillan
Jul 12 '12 at 13:44
$begingroup$
The Sorgenfrey line is totally disconnected. One can prove that an infinite totally disconnected space doesn't have uncountable compact subsets, even when it's not discret.
$endgroup$
– Temitope.A
Jul 20 '12 at 13:08
$begingroup$
@Temitope.A Nonsense. The Cantor set is uncountable, totally disconnected and compact.
$endgroup$
– Henno Brandsma
Jan 28 at 9:14
add a comment |
$begingroup$
Hint: any uncountable set of real numbers contains a strictly increasing infinite sequence.
Hint 2 (added later): show that if a subspace $X$ of the Sorgenfrey line contains a strictly increasing infinite sequence, then $X$ has an open cover with no finite subcover.
answered Jul 12 '12 at 11:58
Colin McQuillanColin McQuillan
4,2681114
$endgroup$
1
$begingroup$
@John But Collin's right.
$endgroup$
– martini
Jul 12 '12 at 12:59
1
$begingroup$
You should try it yourself, for starters. And in the classical topology of $mathbb R$ such sequences can be convergent in contrast to the Sorgenfrey case, where such sequences can't have a convergent subsequence ...
$endgroup$
– martini
Jul 12 '12 at 13:06
$begingroup$
@John: see my second hint above.
$endgroup$
– Colin McQuillan
Jul 12 '12 at 13:44
$begingroup$
The Sorgenfrey line is totally disconnected. One can prove that an infinite totally disconnected space doesn't have uncountable compact subsets, even when it's not discret.
$endgroup$
– Temitope.A
Jul 20 '12 at 13:08
$begingroup$
@Temitope.A Nonsense. The Cantor set is uncountable, totally disconnected and compact.
$endgroup$
– Henno Brandsma
Jan 28 at 9:14
add a comment |
$begingroup$
Hint: any uncountable set of real numbers contains a strictly increasing infinite sequence.
Hint 2 (added later): show that if a subspace $X$ of the Sorgenfrey line contains a strictly increasing infinite sequence, then $X$ has an open cover with no finite subcover.
answered Jul 12 '12 at 11:58
Colin McQuillanColin McQuillan
4,2681114
$endgroup$
Hint: any uncountable set of real numbers contains a strictly increasing infinite sequence.
Hint 2 (added later): show that if a subspace $X$ of the Sorgenfrey line contains a strictly increasing infinite sequence, then $X$ has an open cover with no finite subcover.
answered Jul 12 '12 at 11:58
Colin McQuillanColin McQuillan
4,2681114
edited Jul 12 '12 at 13:43
answered Jul 12 '12 at 11:58
Colin McQuillanColin McQuillan
4,2681114
answered Jul 12 '12 at 11:58
Colin McQuillanColin McQuillan
4,2681114
answered Jul 12 '12 at 11:58
Colin McQuillanColin McQuillan
4,2681114
4,2681114
1
$begingroup$
@John But Collin's right.
$endgroup$
– martini
Jul 12 '12 at 12:59
1
$begingroup$
You should try it yourself, for starters. And in the classical topology of $mathbb R$ such sequences can be convergent in contrast to the Sorgenfrey case, where such sequences can't have a convergent subsequence ...
$endgroup$
– martini
Jul 12 '12 at 13:06
$begingroup$
@John: see my second hint above.
$endgroup$
– Colin McQuillan
Jul 12 '12 at 13:44
$begingroup$
The Sorgenfrey line is totally disconnected. One can prove that an infinite totally disconnected space doesn't have uncountable compact subsets, even when it's not discret.
$endgroup$
– Temitope.A
Jul 20 '12 at 13:08
$begingroup$
@Temitope.A Nonsense. The Cantor set is uncountable, totally disconnected and compact.
$endgroup$
– Henno Brandsma
Jan 28 at 9:14
add a comment |
1
$begingroup$
@John But Collin's right.
$endgroup$
– martini
Jul 12 '12 at 12:59
1
$begingroup$
You should try it yourself, for starters. And in the classical topology of $mathbb R$ such sequences can be convergent in contrast to the Sorgenfrey case, where such sequences can't have a convergent subsequence ...
$endgroup$
– martini
Jul 12 '12 at 13:06
$begingroup$
@John: see my second hint above.
$endgroup$
– Colin McQuillan
Jul 12 '12 at 13:44
$begingroup$
The Sorgenfrey line is totally disconnected. One can prove that an infinite totally disconnected space doesn't have uncountable compact subsets, even when it's not discret.
$endgroup$
– Temitope.A
Jul 20 '12 at 13:08
$begingroup$
@Temitope.A Nonsense. The Cantor set is uncountable, totally disconnected and compact.
$endgroup$
– Henno Brandsma
Jan 28 at 9:14
1
1
$begingroup$
@John But Collin's right.
$endgroup$
– martini
Jul 12 '12 at 12:59
$begingroup$
@John But Collin's right.
$endgroup$
– martini
Jul 12 '12 at 12:59
1
1
$begingroup$
You should try it yourself, for starters. And in the classical topology of $mathbb R$ such sequences can be convergent in contrast to the Sorgenfrey case, where such sequences can't have a convergent subsequence ...
$endgroup$
– martini
Jul 12 '12 at 13:06
$begingroup$
You should try it yourself, for starters. And in the classical topology of $mathbb R$ such sequences can be convergent in contrast to the Sorgenfrey case, where such sequences can't have a convergent subsequence ...
$endgroup$
– martini
Jul 12 '12 at 13:06
$begingroup$
@John: see my second hint above.
$endgroup$
– Colin McQuillan
Jul 12 '12 at 13:44
$begingroup$
@John: see my second hint above.
$endgroup$
– Colin McQuillan
Jul 12 '12 at 13:44
$begingroup$
The Sorgenfrey line is totally disconnected. One can prove that an infinite totally disconnected space doesn't have uncountable compact subsets, even when it's not discret.
$endgroup$
– Temitope.A
Jul 20 '12 at 13:08
$begingroup$
The Sorgenfrey line is totally disconnected. One can prove that an infinite totally disconnected space doesn't have uncountable compact subsets, even when it's not discret.
$endgroup$
– Temitope.A
Jul 20 '12 at 13:08
$begingroup$
@Temitope.A Nonsense. The Cantor set is uncountable, totally disconnected and compact.
$endgroup$
– Henno Brandsma
Jan 28 at 9:14
$begingroup$
@Temitope.A Nonsense. The Cantor set is uncountable, totally disconnected and compact.
$endgroup$
– Henno Brandsma
Jan 28 at 9:14
add a comment |
$begingroup$
not true. The compact subsets of Sorgenfrey line must be finite sets. If it is infinite countable set even it can not be compact in the ususal topology on R
answered Feb 10 '16 at 9:23
cenap ozelcenap ozel
1
$endgroup$
$begingroup$
this doesn't make any sense. [0,1] is compact in $mathbb R$ with the usual topology and is even uncountable infinite. And what has this to do with the Sorgenfrey line anyway??
$endgroup$
– noctusraid
Feb 10 '16 at 9:27
add a comment |
$begingroup$
not true. The compact subsets of Sorgenfrey line must be finite sets. If it is infinite countable set even it can not be compact in the ususal topology on R
answered Feb 10 '16 at 9:23
cenap ozelcenap ozel
1
$endgroup$
$begingroup$
this doesn't make any sense. [0,1] is compact in $mathbb R$ with the usual topology and is even uncountable infinite. And what has this to do with the Sorgenfrey line anyway??
$endgroup$
– noctusraid
Feb 10 '16 at 9:27
add a comment |
$begingroup$
not true. The compact subsets of Sorgenfrey line must be finite sets. If it is infinite countable set even it can not be compact in the ususal topology on R
answered Feb 10 '16 at 9:23
cenap ozelcenap ozel
1
$endgroup$
not true. The compact subsets of Sorgenfrey line must be finite sets. If it is infinite countable set even it can not be compact in the ususal topology on R
answered Feb 10 '16 at 9:23
cenap ozelcenap ozel
1
answered Feb 10 '16 at 9:23
cenap ozelcenap ozel
1
answered Feb 10 '16 at 9:23
cenap ozelcenap ozel
1
answered Feb 10 '16 at 9:23
cenap ozelcenap ozel
1
1
$begingroup$
this doesn't make any sense. [0,1] is compact in $mathbb R$ with the usual topology and is even uncountable infinite. And what has this to do with the Sorgenfrey line anyway??
$endgroup$
– noctusraid
Feb 10 '16 at 9:27
add a comment |
$begingroup$
this doesn't make any sense. [0,1] is compact in $mathbb R$ with the usual topology and is even uncountable infinite. And what has this to do with the Sorgenfrey line anyway??
$endgroup$
– noctusraid
Feb 10 '16 at 9:27
$begingroup$
this doesn't make any sense. [0,1] is compact in $mathbb R$ with the usual topology and is even uncountable infinite. And what has this to do with the Sorgenfrey line anyway??
$endgroup$
– noctusraid
Feb 10 '16 at 9:27
$begingroup$
this doesn't make any sense. [0,1] is compact in $mathbb R$ with the usual topology and is even uncountable infinite. And what has this to do with the Sorgenfrey line anyway??
$endgroup$
– noctusraid
Feb 10 '16 at 9:27
add a comment |
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