$Ucap K$ is compact in the subspace topology
If $X,tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,tau$ then $Ucap K$ is compact.
Attempted proof:
Since $U$ is compact there is a finite sub covering such that $Usubset bigcup_limits{i=1}^{n}B_i$ $B_iin tauforall 1leqslant ileqslant n$
$Ucap Ksubset (bigcup_limits{i=1}^{n}B_i)cap K=bigcup_limits{i=1}^{n}(B_icap K)$
$B_icap K$ is open for the subspace topology for all $i$.
Hence $bigcup_limits{i=1}^{n}(B_icap K)$ is a finite sub covering of $Ucap K$ proving $Ucap K$ to be compact in the subspace topology.
Question:
Is my proof right? If not how should I correct it?
Thanks in advance!
general-topology proof-verification compactness
edited 2 days ago
José Carlos Santos
149k22117219
asked 2 days ago
Pedro Gomes
1,6792720
add a comment |
If $X,tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,tau$ then $Ucap K$ is compact.
Attempted proof:
Since $U$ is compact there is a finite sub covering such that $Usubset bigcup_limits{i=1}^{n}B_i$ $B_iin tauforall 1leqslant ileqslant n$
$Ucap Ksubset (bigcup_limits{i=1}^{n}B_i)cap K=bigcup_limits{i=1}^{n}(B_icap K)$
$B_icap K$ is open for the subspace topology for all $i$.
Hence $bigcup_limits{i=1}^{n}(B_icap K)$ is a finite sub covering of $Ucap K$ proving $Ucap K$ to be compact in the subspace topology.
Question:
Is my proof right? If not how should I correct it?
Thanks in advance!
general-topology proof-verification compactness
edited 2 days ago
José Carlos Santos
149k22117219
asked 2 days ago
Pedro Gomes
1,6792720
I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
– Uncountable
2 days ago
The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
– Uncountable
2 days ago
add a comment |
If $X,tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,tau$ then $Ucap K$ is compact.
Attempted proof:
Since $U$ is compact there is a finite sub covering such that $Usubset bigcup_limits{i=1}^{n}B_i$ $B_iin tauforall 1leqslant ileqslant n$
$Ucap Ksubset (bigcup_limits{i=1}^{n}B_i)cap K=bigcup_limits{i=1}^{n}(B_icap K)$
$B_icap K$ is open for the subspace topology for all $i$.
Hence $bigcup_limits{i=1}^{n}(B_icap K)$ is a finite sub covering of $Ucap K$ proving $Ucap K$ to be compact in the subspace topology.
Question:
Is my proof right? If not how should I correct it?
Thanks in advance!
general-topology proof-verification compactness
edited 2 days ago
José Carlos Santos
149k22117219
asked 2 days ago
Pedro Gomes
1,6792720
If $X,tau$ is a topological space and K is a closed subspace. If $U$ is a compact set of $X,tau$ then $Ucap K$ is compact.
Attempted proof:
Since $U$ is compact there is a finite sub covering such that $Usubset bigcup_limits{i=1}^{n}B_i$ $B_iin tauforall 1leqslant ileqslant n$
$Ucap Ksubset (bigcup_limits{i=1}^{n}B_i)cap K=bigcup_limits{i=1}^{n}(B_icap K)$
$B_icap K$ is open for the subspace topology for all $i$.
Hence $bigcup_limits{i=1}^{n}(B_icap K)$ is a finite sub covering of $Ucap K$ proving $Ucap K$ to be compact in the subspace topology.
Question:
Is my proof right? If not how should I correct it?
Thanks in advance!
general-topology proof-verification compactness
general-topology proof-verification compactness
edited 2 days ago
José Carlos Santos
149k22117219
asked 2 days ago
Pedro Gomes
1,6792720
edited 2 days ago
José Carlos Santos
149k22117219
asked 2 days ago
Pedro Gomes
1,6792720
edited 2 days ago
José Carlos Santos
149k22117219
edited 2 days ago
José Carlos Santos
149k22117219
edited 2 days ago
José Carlos Santos
149k22117219
149k22117219
asked 2 days ago
Pedro Gomes
1,6792720
asked 2 days ago
Pedro Gomes
1,6792720
asked 2 days ago
Pedro Gomes
1,6792720
1,6792720
I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
– Uncountable
2 days ago
The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
– Uncountable
2 days ago
add a comment |
I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
– Uncountable
2 days ago
The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
– Uncountable
2 days ago
I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
– Uncountable
2 days ago
I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
– Uncountable
2 days ago
The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
– Uncountable
2 days ago
The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
– Uncountable
2 days ago
add a comment |
1 Answer
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The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.
Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.
answered 2 days ago
José Carlos Santos
149k22117219
add a comment |
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1 Answer
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oldest
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votes
The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.
Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.
answered 2 days ago
José Carlos Santos
149k22117219
add a comment |
The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.
Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.
answered 2 days ago
José Carlos Santos
149k22117219
add a comment |
The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.
Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.
answered 2 days ago
José Carlos Santos
149k22117219
The proof is wrong from the start. Every topological space has a finite covering: just take the covering with a single open set, which is the whole space.
Let $(O_lambda)_{lambdainLambda}$ be an open cover of $Ucap K$. Add to this cover the set $Ucap K^complement$ (which is an open subset of $U$). Then you have an open cover of $U$. Since $U$ is compact, it has a finite subcover. The elements of this finite subcover which are distinct from $Ucap K^complement$ form a finite subcover of $Ucap K$.
answered 2 days ago
José Carlos Santos
149k22117219
edited 2 days ago
answered 2 days ago
José Carlos Santos
149k22117219
answered 2 days ago
José Carlos Santos
149k22117219
answered 2 days ago
José Carlos Santos
149k22117219
149k22117219
add a comment |
add a comment |
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I think you're misusing the definition of compactness: it says that every open cover has a finite subcover (not just that a finite cover exists).
– Uncountable
2 days ago
The statement is equivalent to: a closed subspace A of a compact space B is compact. You can prove this by constructing an open cover of B from an open cover of A (which is given by open subsets of B intersected with A) by adding the complement of A, and extracting a finite subcover.
– Uncountable
2 days ago