Is an open subset of an open set of a topological space open?
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If $X$ is an topological space and $ U subseteq V subseteq X$ with $V$ open in $X$ and $U$ open in $V$, is it true that $U$ is open in $X$?
general-topology
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add a comment |
$begingroup$
If $X$ is an topological space and $ U subseteq V subseteq X$ with $V$ open in $X$ and $U$ open in $V$, is it true that $U$ is open in $X$?
general-topology
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add a comment |
$begingroup$
If $X$ is an topological space and $ U subseteq V subseteq X$ with $V$ open in $X$ and $U$ open in $V$, is it true that $U$ is open in $X$?
general-topology
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If $X$ is an topological space and $ U subseteq V subseteq X$ with $V$ open in $X$ and $U$ open in $V$, is it true that $U$ is open in $X$?
general-topology
general-topology
edited Dec 5 '18 at 16:57
Andrews
3901317
3901317
asked Dec 5 '18 at 16:07
gengen
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4532521
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2 Answers
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Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.
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Thank you (five more characters)
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– gen
Dec 5 '18 at 16:11
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Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.
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Is this just the definition of U being open in V?
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– gen
Dec 5 '18 at 16:13
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Yes. This is how a subspace topology is defined.
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– Thomas Shelby
Dec 5 '18 at 16:15
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.
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$begingroup$
Thank you (five more characters)
$endgroup$
– gen
Dec 5 '18 at 16:11
add a comment |
$begingroup$
Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.
$endgroup$
$begingroup$
Thank you (five more characters)
$endgroup$
– gen
Dec 5 '18 at 16:11
add a comment |
$begingroup$
Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.
$endgroup$
Yes. $U$ open in $V$ means that $U=Vcap W$ for some $W$ open in $X$. So $U$ is an intersection of two open sets in $X$.
answered Dec 5 '18 at 16:10
ajotatxeajotatxe
53.8k23890
53.8k23890
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Thank you (five more characters)
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– gen
Dec 5 '18 at 16:11
add a comment |
$begingroup$
Thank you (five more characters)
$endgroup$
– gen
Dec 5 '18 at 16:11
$begingroup$
Thank you (five more characters)
$endgroup$
– gen
Dec 5 '18 at 16:11
$begingroup$
Thank you (five more characters)
$endgroup$
– gen
Dec 5 '18 at 16:11
add a comment |
$begingroup$
Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.
$endgroup$
$begingroup$
Is this just the definition of U being open in V?
$endgroup$
– gen
Dec 5 '18 at 16:13
$begingroup$
Yes. This is how a subspace topology is defined.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 16:15
add a comment |
$begingroup$
Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.
$endgroup$
$begingroup$
Is this just the definition of U being open in V?
$endgroup$
– gen
Dec 5 '18 at 16:13
$begingroup$
Yes. This is how a subspace topology is defined.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 16:15
add a comment |
$begingroup$
Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.
$endgroup$
Hint: $U $ is open in $V $ iff $U=Vcap S $ for some $S $ open in $X $.
edited Dec 5 '18 at 16:23
answered Dec 5 '18 at 16:12
Thomas ShelbyThomas Shelby
2,710421
2,710421
$begingroup$
Is this just the definition of U being open in V?
$endgroup$
– gen
Dec 5 '18 at 16:13
$begingroup$
Yes. This is how a subspace topology is defined.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 16:15
add a comment |
$begingroup$
Is this just the definition of U being open in V?
$endgroup$
– gen
Dec 5 '18 at 16:13
$begingroup$
Yes. This is how a subspace topology is defined.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 16:15
$begingroup$
Is this just the definition of U being open in V?
$endgroup$
– gen
Dec 5 '18 at 16:13
$begingroup$
Is this just the definition of U being open in V?
$endgroup$
– gen
Dec 5 '18 at 16:13
$begingroup$
Yes. This is how a subspace topology is defined.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 16:15
$begingroup$
Yes. This is how a subspace topology is defined.
$endgroup$
– Thomas Shelby
Dec 5 '18 at 16:15
add a comment |
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