Intersection of sequentially compact sets
$begingroup$
Let A, B (non-empty) sequentially compact sets.
Then the intersection of A and B is sequentially compact.
One can prove this hypothesis by selecting a sequence in the intersection and observing a convergent subsequence in A the limit of which is a. Then there exists an subsubsequence in B and as the limits coincide and the hypothesis is proven.
Now, I am wondering, if the following is a valid proof (or it is just a reformulation of the argument in proof 1):
Again, select a sequence in the intersection and observe a convergent subsequence. As the sequence and thus the convergent subsequence with it's limit are in (say) A, the limit must be in B, too, otherwise there are elements in the sequence which are not in B, which is a contradiction to the givens. Shouldn't that proof the hypothesis as well?
real-analysis sequences-and-series general-topology compactness
asked Dec 6 '18 at 20:57
QuantaurixQuantaurix
132
$endgroup$
add a comment |
$begingroup$
Let A, B (non-empty) sequentially compact sets.
Then the intersection of A and B is sequentially compact.
One can prove this hypothesis by selecting a sequence in the intersection and observing a convergent subsequence in A the limit of which is a. Then there exists an subsubsequence in B and as the limits coincide and the hypothesis is proven.
Now, I am wondering, if the following is a valid proof (or it is just a reformulation of the argument in proof 1):
Again, select a sequence in the intersection and observe a convergent subsequence. As the sequence and thus the convergent subsequence with it's limit are in (say) A, the limit must be in B, too, otherwise there are elements in the sequence which are not in B, which is a contradiction to the givens. Shouldn't that proof the hypothesis as well?
real-analysis sequences-and-series general-topology compactness
asked Dec 6 '18 at 20:57
QuantaurixQuantaurix
132
$endgroup$
add a comment |
$begingroup$
Let A, B (non-empty) sequentially compact sets.
Then the intersection of A and B is sequentially compact.
One can prove this hypothesis by selecting a sequence in the intersection and observing a convergent subsequence in A the limit of which is a. Then there exists an subsubsequence in B and as the limits coincide and the hypothesis is proven.
Now, I am wondering, if the following is a valid proof (or it is just a reformulation of the argument in proof 1):
Again, select a sequence in the intersection and observe a convergent subsequence. As the sequence and thus the convergent subsequence with it's limit are in (say) A, the limit must be in B, too, otherwise there are elements in the sequence which are not in B, which is a contradiction to the givens. Shouldn't that proof the hypothesis as well?
real-analysis sequences-and-series general-topology compactness
asked Dec 6 '18 at 20:57
QuantaurixQuantaurix
132
$endgroup$
Let A, B (non-empty) sequentially compact sets.
Then the intersection of A and B is sequentially compact.
One can prove this hypothesis by selecting a sequence in the intersection and observing a convergent subsequence in A the limit of which is a. Then there exists an subsubsequence in B and as the limits coincide and the hypothesis is proven.
Now, I am wondering, if the following is a valid proof (or it is just a reformulation of the argument in proof 1):
Again, select a sequence in the intersection and observe a convergent subsequence. As the sequence and thus the convergent subsequence with it's limit are in (say) A, the limit must be in B, too, otherwise there are elements in the sequence which are not in B, which is a contradiction to the givens. Shouldn't that proof the hypothesis as well?
real-analysis sequences-and-series general-topology compactness
real-analysis sequences-and-series general-topology compactness
asked Dec 6 '18 at 20:57
QuantaurixQuantaurix
132
asked Dec 6 '18 at 20:57
QuantaurixQuantaurix
132
asked Dec 6 '18 at 20:57
QuantaurixQuantaurix
132
asked Dec 6 '18 at 20:57
QuantaurixQuantaurix
132
asked Dec 6 '18 at 20:57
QuantaurixQuantaurix
132
132
add a comment |
add a comment |
1 Answer
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$begingroup$
The answer is No if you are working in a general topological space. If the condition is restricted to a metric space then both your sequentially compact subsets become closed by being compact spaces in the metric space. Then your argument should work.
What you are looking for should be a sequentially compact subset $B$ in some first-countable topological space which is not closed, thus the limit of some sequence $(x_n)$ is not contained in $B$. Choose another sequentially compact subset $A$ which is closed contains $(x_n)$, then although $(x_n)$ contains itself as a subsequence, the limit is not in $B$.
See the top answer on this post:
Sequentially compact subset of Hausdorff space is Closed?
answered Dec 6 '18 at 21:38
William SunWilliam Sun
471111
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The answer is No if you are working in a general topological space. If the condition is restricted to a metric space then both your sequentially compact subsets become closed by being compact spaces in the metric space. Then your argument should work.
What you are looking for should be a sequentially compact subset $B$ in some first-countable topological space which is not closed, thus the limit of some sequence $(x_n)$ is not contained in $B$. Choose another sequentially compact subset $A$ which is closed contains $(x_n)$, then although $(x_n)$ contains itself as a subsequence, the limit is not in $B$.
See the top answer on this post:
Sequentially compact subset of Hausdorff space is Closed?
answered Dec 6 '18 at 21:38
William SunWilliam Sun
471111
$endgroup$
add a comment |
$begingroup$
The answer is No if you are working in a general topological space. If the condition is restricted to a metric space then both your sequentially compact subsets become closed by being compact spaces in the metric space. Then your argument should work.
What you are looking for should be a sequentially compact subset $B$ in some first-countable topological space which is not closed, thus the limit of some sequence $(x_n)$ is not contained in $B$. Choose another sequentially compact subset $A$ which is closed contains $(x_n)$, then although $(x_n)$ contains itself as a subsequence, the limit is not in $B$.
See the top answer on this post:
Sequentially compact subset of Hausdorff space is Closed?
answered Dec 6 '18 at 21:38
William SunWilliam Sun
471111
$endgroup$
add a comment |
$begingroup$
The answer is No if you are working in a general topological space. If the condition is restricted to a metric space then both your sequentially compact subsets become closed by being compact spaces in the metric space. Then your argument should work.
What you are looking for should be a sequentially compact subset $B$ in some first-countable topological space which is not closed, thus the limit of some sequence $(x_n)$ is not contained in $B$. Choose another sequentially compact subset $A$ which is closed contains $(x_n)$, then although $(x_n)$ contains itself as a subsequence, the limit is not in $B$.
See the top answer on this post:
Sequentially compact subset of Hausdorff space is Closed?
answered Dec 6 '18 at 21:38
William SunWilliam Sun
471111
$endgroup$
The answer is No if you are working in a general topological space. If the condition is restricted to a metric space then both your sequentially compact subsets become closed by being compact spaces in the metric space. Then your argument should work.
What you are looking for should be a sequentially compact subset $B$ in some first-countable topological space which is not closed, thus the limit of some sequence $(x_n)$ is not contained in $B$. Choose another sequentially compact subset $A$ which is closed contains $(x_n)$, then although $(x_n)$ contains itself as a subsequence, the limit is not in $B$.
See the top answer on this post:
Sequentially compact subset of Hausdorff space is Closed?
answered Dec 6 '18 at 21:38
William SunWilliam Sun
471111
edited Dec 6 '18 at 22:22
answered Dec 6 '18 at 21:38
William SunWilliam Sun
471111
answered Dec 6 '18 at 21:38
William SunWilliam Sun
471111
answered Dec 6 '18 at 21:38
William SunWilliam Sun
471111
471111
add a comment |
add a comment |
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