A question about $C^{*}$-embedded and $C$-embedded
$begingroup$
A subspace $S$ of $X$ is $C$-embedded in $X$ if every function in $C(S)$ can be extended to a function in $C(X)$. A subspace $S$ of $X$ is $C^{*}$-embedded in $X$ if every function in $C^{*}(S)$ can be extended to a function in $C^{*}(X)$.
$C^{*}(X)= { f in C(X) | f quad is quad bounded }$
Theorem 1: A subspace $S$ of $X$ is $C^{*}$-embedded in $X$ if only if any two completely separated sets in $S$ are completely separated sets in $X$.
Theorem 2: A $C^{*}$-embedded is $C$-embedded if only if it is completely separated from every zero-set disjoint from it.
According to the two above-mentioned theorems can be shown below
problem? can you help me?
The following are equivalent for any Hausdorff space $X$.
1: $X$ is normal.
2:Any two disjoint closed sets are completely separated.
3:Every closed set is $C^{*}$-embedded.
4:Every closed set is $C$-embedded.
general-topology topological-groups
$endgroup$
add a comment |
$begingroup$
A subspace $S$ of $X$ is $C$-embedded in $X$ if every function in $C(S)$ can be extended to a function in $C(X)$. A subspace $S$ of $X$ is $C^{*}$-embedded in $X$ if every function in $C^{*}(S)$ can be extended to a function in $C^{*}(X)$.
$C^{*}(X)= { f in C(X) | f quad is quad bounded }$
Theorem 1: A subspace $S$ of $X$ is $C^{*}$-embedded in $X$ if only if any two completely separated sets in $S$ are completely separated sets in $X$.
Theorem 2: A $C^{*}$-embedded is $C$-embedded if only if it is completely separated from every zero-set disjoint from it.
According to the two above-mentioned theorems can be shown below
problem? can you help me?
The following are equivalent for any Hausdorff space $X$.
1: $X$ is normal.
2:Any two disjoint closed sets are completely separated.
3:Every closed set is $C^{*}$-embedded.
4:Every closed set is $C$-embedded.
general-topology topological-groups
$endgroup$
1
$begingroup$
1 implies 2 is Urysohn's lemma.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 23:12
add a comment |
$begingroup$
A subspace $S$ of $X$ is $C$-embedded in $X$ if every function in $C(S)$ can be extended to a function in $C(X)$. A subspace $S$ of $X$ is $C^{*}$-embedded in $X$ if every function in $C^{*}(S)$ can be extended to a function in $C^{*}(X)$.
$C^{*}(X)= { f in C(X) | f quad is quad bounded }$
Theorem 1: A subspace $S$ of $X$ is $C^{*}$-embedded in $X$ if only if any two completely separated sets in $S$ are completely separated sets in $X$.
Theorem 2: A $C^{*}$-embedded is $C$-embedded if only if it is completely separated from every zero-set disjoint from it.
According to the two above-mentioned theorems can be shown below
problem? can you help me?
The following are equivalent for any Hausdorff space $X$.
1: $X$ is normal.
2:Any two disjoint closed sets are completely separated.
3:Every closed set is $C^{*}$-embedded.
4:Every closed set is $C$-embedded.
general-topology topological-groups
$endgroup$
A subspace $S$ of $X$ is $C$-embedded in $X$ if every function in $C(S)$ can be extended to a function in $C(X)$. A subspace $S$ of $X$ is $C^{*}$-embedded in $X$ if every function in $C^{*}(S)$ can be extended to a function in $C^{*}(X)$.
$C^{*}(X)= { f in C(X) | f quad is quad bounded }$
Theorem 1: A subspace $S$ of $X$ is $C^{*}$-embedded in $X$ if only if any two completely separated sets in $S$ are completely separated sets in $X$.
Theorem 2: A $C^{*}$-embedded is $C$-embedded if only if it is completely separated from every zero-set disjoint from it.
According to the two above-mentioned theorems can be shown below
problem? can you help me?
The following are equivalent for any Hausdorff space $X$.
1: $X$ is normal.
2:Any two disjoint closed sets are completely separated.
3:Every closed set is $C^{*}$-embedded.
4:Every closed set is $C$-embedded.
general-topology topological-groups
general-topology topological-groups
asked Dec 26 '18 at 19:18
user387219
1
$begingroup$
1 implies 2 is Urysohn's lemma.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 23:12
add a comment |
1
$begingroup$
1 implies 2 is Urysohn's lemma.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 23:12
1
1
$begingroup$
1 implies 2 is Urysohn's lemma.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 23:12
$begingroup$
1 implies 2 is Urysohn's lemma.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 23:12
add a comment |
1 Answer
1
active
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$begingroup$
Conditions 1-4 are equivalent. Implications $4Rightarrow 3$, $3Rightarrow 2$, and $2Rightarrow 1$ are obvious. Implication $1Rightarrow 2$ follows from Urysohn’s lemma [Eng, 1.5.11 and p.42]. $1Rightarrow 3,4 $ by the Tietze-Urysohn Theorem [Eng, 2.1.8]. $2Rightarrow 3$ also follows from Theorem 1, and $3Rightarrow 4$ from Theorem 2.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 3 at 4:12
Alex RavskyAlex Ravsky
42.6k32383
$endgroup$
add a comment |
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$begingroup$
Conditions 1-4 are equivalent. Implications $4Rightarrow 3$, $3Rightarrow 2$, and $2Rightarrow 1$ are obvious. Implication $1Rightarrow 2$ follows from Urysohn’s lemma [Eng, 1.5.11 and p.42]. $1Rightarrow 3,4 $ by the Tietze-Urysohn Theorem [Eng, 2.1.8]. $2Rightarrow 3$ also follows from Theorem 1, and $3Rightarrow 4$ from Theorem 2.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 3 at 4:12
Alex RavskyAlex Ravsky
42.6k32383
$endgroup$
add a comment |
$begingroup$
Conditions 1-4 are equivalent. Implications $4Rightarrow 3$, $3Rightarrow 2$, and $2Rightarrow 1$ are obvious. Implication $1Rightarrow 2$ follows from Urysohn’s lemma [Eng, 1.5.11 and p.42]. $1Rightarrow 3,4 $ by the Tietze-Urysohn Theorem [Eng, 2.1.8]. $2Rightarrow 3$ also follows from Theorem 1, and $3Rightarrow 4$ from Theorem 2.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 3 at 4:12
Alex RavskyAlex Ravsky
42.6k32383
$endgroup$
add a comment |
$begingroup$
Conditions 1-4 are equivalent. Implications $4Rightarrow 3$, $3Rightarrow 2$, and $2Rightarrow 1$ are obvious. Implication $1Rightarrow 2$ follows from Urysohn’s lemma [Eng, 1.5.11 and p.42]. $1Rightarrow 3,4 $ by the Tietze-Urysohn Theorem [Eng, 2.1.8]. $2Rightarrow 3$ also follows from Theorem 1, and $3Rightarrow 4$ from Theorem 2.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 3 at 4:12
Alex RavskyAlex Ravsky
42.6k32383
$endgroup$
Conditions 1-4 are equivalent. Implications $4Rightarrow 3$, $3Rightarrow 2$, and $2Rightarrow 1$ are obvious. Implication $1Rightarrow 2$ follows from Urysohn’s lemma [Eng, 1.5.11 and p.42]. $1Rightarrow 3,4 $ by the Tietze-Urysohn Theorem [Eng, 2.1.8]. $2Rightarrow 3$ also follows from Theorem 1, and $3Rightarrow 4$ from Theorem 2.
References
[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
answered Jan 3 at 4:12
Alex RavskyAlex Ravsky
42.6k32383
answered Jan 3 at 4:12
Alex RavskyAlex Ravsky
42.6k32383
answered Jan 3 at 4:12
Alex RavskyAlex Ravsky
42.6k32383
answered Jan 3 at 4:12
Alex RavskyAlex Ravsky
42.6k32383
42.6k32383
add a comment |
add a comment |
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1 implies 2 is Urysohn's lemma.
$endgroup$
– Henno Brandsma
Dec 26 '18 at 23:12