Does paracompact Hausdorff imply perfectly normal?
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That paracompact Hausdorff implies normal is standard and there are examples on StackExchange of perfectly normal Hausdorff spaces that are not paracompact, but I'm not sure of the answer, especially since paracompact spaces are collection-wise normal and the latter is not related to perfectly normal. (The standard example of a collection-wise normal space that is not perfectly normal is $omega_1$, which is not paracompact.)
general-topology examples-counterexamples separation-axioms paracompactness
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That paracompact Hausdorff implies normal is standard and there are examples on StackExchange of perfectly normal Hausdorff spaces that are not paracompact, but I'm not sure of the answer, especially since paracompact spaces are collection-wise normal and the latter is not related to perfectly normal. (The standard example of a collection-wise normal space that is not perfectly normal is $omega_1$, which is not paracompact.)
general-topology examples-counterexamples separation-axioms paracompactness
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add a comment |
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That paracompact Hausdorff implies normal is standard and there are examples on StackExchange of perfectly normal Hausdorff spaces that are not paracompact, but I'm not sure of the answer, especially since paracompact spaces are collection-wise normal and the latter is not related to perfectly normal. (The standard example of a collection-wise normal space that is not perfectly normal is $omega_1$, which is not paracompact.)
general-topology examples-counterexamples separation-axioms paracompactness
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That paracompact Hausdorff implies normal is standard and there are examples on StackExchange of perfectly normal Hausdorff spaces that are not paracompact, but I'm not sure of the answer, especially since paracompact spaces are collection-wise normal and the latter is not related to perfectly normal. (The standard example of a collection-wise normal space that is not perfectly normal is $omega_1$, which is not paracompact.)
general-topology examples-counterexamples separation-axioms paracompactness
general-topology examples-counterexamples separation-axioms paracompactness
edited Dec 9 '18 at 3:29
Eric Wofsey
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asked Jun 24 '14 at 10:16
chrystomathchrystomath
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No. Consider the closed ordinal space $omega_1+1 = [0,omega_1]$. Since it is compact, it is clearly paracompact, and is also Hausdorff. But it is not perfectly normal: the (closed) singleton ${ omega_1 }$ is not a Gδ subset.
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1
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Thanks! So even a compact hausdorff space need not be perfectly normal.
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– chrystomath
Jun 24 '14 at 10:51
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@chrystomath: You're welcome! Yes, compact Hausdorff spaces need not be perfectly normal. In fact, they may fail to be even hereditarily normal: for example, the Tychonoff plank.
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– user642796
Jun 24 '14 at 11:31
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1 Answer
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1 Answer
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No. Consider the closed ordinal space $omega_1+1 = [0,omega_1]$. Since it is compact, it is clearly paracompact, and is also Hausdorff. But it is not perfectly normal: the (closed) singleton ${ omega_1 }$ is not a Gδ subset.
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1
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Thanks! So even a compact hausdorff space need not be perfectly normal.
$endgroup$
– chrystomath
Jun 24 '14 at 10:51
1
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@chrystomath: You're welcome! Yes, compact Hausdorff spaces need not be perfectly normal. In fact, they may fail to be even hereditarily normal: for example, the Tychonoff plank.
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– user642796
Jun 24 '14 at 11:31
add a comment |
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No. Consider the closed ordinal space $omega_1+1 = [0,omega_1]$. Since it is compact, it is clearly paracompact, and is also Hausdorff. But it is not perfectly normal: the (closed) singleton ${ omega_1 }$ is not a Gδ subset.
$endgroup$
1
$begingroup$
Thanks! So even a compact hausdorff space need not be perfectly normal.
$endgroup$
– chrystomath
Jun 24 '14 at 10:51
1
$begingroup$
@chrystomath: You're welcome! Yes, compact Hausdorff spaces need not be perfectly normal. In fact, they may fail to be even hereditarily normal: for example, the Tychonoff plank.
$endgroup$
– user642796
Jun 24 '14 at 11:31
add a comment |
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No. Consider the closed ordinal space $omega_1+1 = [0,omega_1]$. Since it is compact, it is clearly paracompact, and is also Hausdorff. But it is not perfectly normal: the (closed) singleton ${ omega_1 }$ is not a Gδ subset.
$endgroup$
No. Consider the closed ordinal space $omega_1+1 = [0,omega_1]$. Since it is compact, it is clearly paracompact, and is also Hausdorff. But it is not perfectly normal: the (closed) singleton ${ omega_1 }$ is not a Gδ subset.
answered Jun 24 '14 at 10:22
user642796user642796
44.5k560116
44.5k560116
1
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Thanks! So even a compact hausdorff space need not be perfectly normal.
$endgroup$
– chrystomath
Jun 24 '14 at 10:51
1
$begingroup$
@chrystomath: You're welcome! Yes, compact Hausdorff spaces need not be perfectly normal. In fact, they may fail to be even hereditarily normal: for example, the Tychonoff plank.
$endgroup$
– user642796
Jun 24 '14 at 11:31
add a comment |
1
$begingroup$
Thanks! So even a compact hausdorff space need not be perfectly normal.
$endgroup$
– chrystomath
Jun 24 '14 at 10:51
1
$begingroup$
@chrystomath: You're welcome! Yes, compact Hausdorff spaces need not be perfectly normal. In fact, they may fail to be even hereditarily normal: for example, the Tychonoff plank.
$endgroup$
– user642796
Jun 24 '14 at 11:31
1
1
$begingroup$
Thanks! So even a compact hausdorff space need not be perfectly normal.
$endgroup$
– chrystomath
Jun 24 '14 at 10:51
$begingroup$
Thanks! So even a compact hausdorff space need not be perfectly normal.
$endgroup$
– chrystomath
Jun 24 '14 at 10:51
1
1
$begingroup$
@chrystomath: You're welcome! Yes, compact Hausdorff spaces need not be perfectly normal. In fact, they may fail to be even hereditarily normal: for example, the Tychonoff plank.
$endgroup$
– user642796
Jun 24 '14 at 11:31
$begingroup$
@chrystomath: You're welcome! Yes, compact Hausdorff spaces need not be perfectly normal. In fact, they may fail to be even hereditarily normal: for example, the Tychonoff plank.
$endgroup$
– user642796
Jun 24 '14 at 11:31
add a comment |
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