Mixing
Show the closure of a subset of a complete metric space is compact.
$begingroup$
"Let $A$ be a subset of a complete metric space. Assume that for all $ε > 0$, there exists a compact set $A_ε$ so that $∀ x ∈ A, d(x, A_ε)<ε$. Show that $A$'s closure is compact."
I am trying to prove it with "Cauchy sequences in complete metric space are convergent" and "sequential compact set is compact". But I do not know how to prove all sequence in A(or A's closure?) has a cauchy subsequence.
Any help or advice would be greatly appreciated. Thank you in advance!
real-analysis general-topology metric-spaces
edited Jan 18 at 6:05
Sou
3,1842823
asked Jan 18 at 6:01
Eric JEric J
163
$endgroup$
add a comment |
$begingroup$
"Let $A$ be a subset of a complete metric space. Assume that for all $ε > 0$, there exists a compact set $A_ε$ so that $∀ x ∈ A, d(x, A_ε)<ε$. Show that $A$'s closure is compact."
I am trying to prove it with "Cauchy sequences in complete metric space are convergent" and "sequential compact set is compact". But I do not know how to prove all sequence in A(or A's closure?) has a cauchy subsequence.
Any help or advice would be greatly appreciated. Thank you in advance!
real-analysis general-topology metric-spaces
edited Jan 18 at 6:05
Sou
3,1842823
asked Jan 18 at 6:01
Eric JEric J
163
$endgroup$
$begingroup$
I would suggest taking the intersection of all sets $A_varepsilon$. At first glance, this looks like the closure of $A$.
$endgroup$
– Lubin
Jan 18 at 6:07
$begingroup$
@Lubin why would any finite subfamily of these $A_varepsilon$ intersect?
$endgroup$
– Henno Brandsma
Jan 18 at 9:20
$begingroup$
@HennoBrandsma, it seems, after a second glance, that my suggestion is worthless. Thanks for encouraging me to think.
$endgroup$
– Lubin
Jan 18 at 14:51
add a comment |
$begingroup$
"Let $A$ be a subset of a complete metric space. Assume that for all $ε > 0$, there exists a compact set $A_ε$ so that $∀ x ∈ A, d(x, A_ε)<ε$. Show that $A$'s closure is compact."
I am trying to prove it with "Cauchy sequences in complete metric space are convergent" and "sequential compact set is compact". But I do not know how to prove all sequence in A(or A's closure?) has a cauchy subsequence.
Any help or advice would be greatly appreciated. Thank you in advance!
real-analysis general-topology metric-spaces
edited Jan 18 at 6:05
Sou
3,1842823
asked Jan 18 at 6:01
Eric JEric J
163
$endgroup$
"Let $A$ be a subset of a complete metric space. Assume that for all $ε > 0$, there exists a compact set $A_ε$ so that $∀ x ∈ A, d(x, A_ε)<ε$. Show that $A$'s closure is compact."
I am trying to prove it with "Cauchy sequences in complete metric space are convergent" and "sequential compact set is compact". But I do not know how to prove all sequence in A(or A's closure?) has a cauchy subsequence.
Any help or advice would be greatly appreciated. Thank you in advance!
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
edited Jan 18 at 6:05
Sou
3,1842823
asked Jan 18 at 6:01
Eric JEric J
163
edited Jan 18 at 6:05
Sou
3,1842823
asked Jan 18 at 6:01
Eric JEric J
163
edited Jan 18 at 6:05
Sou
3,1842823
edited Jan 18 at 6:05
Sou
3,1842823
edited Jan 18 at 6:05
Sou
3,1842823
3,1842823
asked Jan 18 at 6:01
Eric JEric J
163
asked Jan 18 at 6:01
Eric JEric J
163
asked Jan 18 at 6:01
Eric JEric J
163
163
$begingroup$
I would suggest taking the intersection of all sets $A_varepsilon$. At first glance, this looks like the closure of $A$.
$endgroup$
– Lubin
Jan 18 at 6:07
$begingroup$
@Lubin why would any finite subfamily of these $A_varepsilon$ intersect?
$endgroup$
– Henno Brandsma
Jan 18 at 9:20
$begingroup$
@HennoBrandsma, it seems, after a second glance, that my suggestion is worthless. Thanks for encouraging me to think.
$endgroup$
– Lubin
Jan 18 at 14:51
add a comment |
$begingroup$
I would suggest taking the intersection of all sets $A_varepsilon$. At first glance, this looks like the closure of $A$.
$endgroup$
– Lubin
Jan 18 at 6:07
$begingroup$
@Lubin why would any finite subfamily of these $A_varepsilon$ intersect?
$endgroup$
– Henno Brandsma
Jan 18 at 9:20
$begingroup$
@HennoBrandsma, it seems, after a second glance, that my suggestion is worthless. Thanks for encouraging me to think.
$endgroup$
– Lubin
Jan 18 at 14:51
$begingroup$
I would suggest taking the intersection of all sets $A_varepsilon$. At first glance, this looks like the closure of $A$.
$endgroup$
– Lubin
Jan 18 at 6:07
$begingroup$
I would suggest taking the intersection of all sets $A_varepsilon$. At first glance, this looks like the closure of $A$.
$endgroup$
– Lubin
Jan 18 at 6:07
$begingroup$
@Lubin why would any finite subfamily of these $A_varepsilon$ intersect?
$endgroup$
– Henno Brandsma
Jan 18 at 9:20
$begingroup$
@Lubin why would any finite subfamily of these $A_varepsilon$ intersect?
$endgroup$
– Henno Brandsma
Jan 18 at 9:20
$begingroup$
@HennoBrandsma, it seems, after a second glance, that my suggestion is worthless. Thanks for encouraging me to think.
$endgroup$
– Lubin
Jan 18 at 14:51
$begingroup$
@HennoBrandsma, it seems, after a second glance, that my suggestion is worthless. Thanks for encouraging me to think.
$endgroup$
– Lubin
Jan 18 at 14:51
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I think the best proof is to use total boundedness. Let $epsilon >0$. Then $A_{epsilon /2}$ is compact, hence totally bounded. You can cover it by a finite number of open balls of radius $epsilon /2$. Call these balls $B(x_i,epsilon /2), 1leq i leq n$. Then verify that $A$ is covered by the balls $B(x_i,epsilon ), 1leq i leq n$. It follows that $A$ is totally bounded. Since $X$ is complete it follows that the closure of $A$ is compact. [ My comment below may be useful here].
answered Jan 18 at 6:22
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
One also needs to remark that the closure of a totally bounded set is still totally bounded.
$endgroup$
– Henno Brandsma
Jan 18 at 9:22
$begingroup$
@HennoBrandsma I am using a theorem which says that the closure of a subset $A$ of a complete metric space is compact iff $A$ is totally bounded. Yes, the proof of this uses the fact that closure of totally bounded set is totally bounded.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the best proof is to use total boundedness. Let $epsilon >0$. Then $A_{epsilon /2}$ is compact, hence totally bounded. You can cover it by a finite number of open balls of radius $epsilon /2$. Call these balls $B(x_i,epsilon /2), 1leq i leq n$. Then verify that $A$ is covered by the balls $B(x_i,epsilon ), 1leq i leq n$. It follows that $A$ is totally bounded. Since $X$ is complete it follows that the closure of $A$ is compact. [ My comment below may be useful here].
answered Jan 18 at 6:22
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
One also needs to remark that the closure of a totally bounded set is still totally bounded.
$endgroup$
– Henno Brandsma
Jan 18 at 9:22
$begingroup$
@HennoBrandsma I am using a theorem which says that the closure of a subset $A$ of a complete metric space is compact iff $A$ is totally bounded. Yes, the proof of this uses the fact that closure of totally bounded set is totally bounded.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:28
add a comment |
$begingroup$
I think the best proof is to use total boundedness. Let $epsilon >0$. Then $A_{epsilon /2}$ is compact, hence totally bounded. You can cover it by a finite number of open balls of radius $epsilon /2$. Call these balls $B(x_i,epsilon /2), 1leq i leq n$. Then verify that $A$ is covered by the balls $B(x_i,epsilon ), 1leq i leq n$. It follows that $A$ is totally bounded. Since $X$ is complete it follows that the closure of $A$ is compact. [ My comment below may be useful here].
answered Jan 18 at 6:22
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
$begingroup$
One also needs to remark that the closure of a totally bounded set is still totally bounded.
$endgroup$
– Henno Brandsma
Jan 18 at 9:22
$begingroup$
@HennoBrandsma I am using a theorem which says that the closure of a subset $A$ of a complete metric space is compact iff $A$ is totally bounded. Yes, the proof of this uses the fact that closure of totally bounded set is totally bounded.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:28
add a comment |
$begingroup$
I think the best proof is to use total boundedness. Let $epsilon >0$. Then $A_{epsilon /2}$ is compact, hence totally bounded. You can cover it by a finite number of open balls of radius $epsilon /2$. Call these balls $B(x_i,epsilon /2), 1leq i leq n$. Then verify that $A$ is covered by the balls $B(x_i,epsilon ), 1leq i leq n$. It follows that $A$ is totally bounded. Since $X$ is complete it follows that the closure of $A$ is compact. [ My comment below may be useful here].
answered Jan 18 at 6:22
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
$endgroup$
I think the best proof is to use total boundedness. Let $epsilon >0$. Then $A_{epsilon /2}$ is compact, hence totally bounded. You can cover it by a finite number of open balls of radius $epsilon /2$. Call these balls $B(x_i,epsilon /2), 1leq i leq n$. Then verify that $A$ is covered by the balls $B(x_i,epsilon ), 1leq i leq n$. It follows that $A$ is totally bounded. Since $X$ is complete it follows that the closure of $A$ is compact. [ My comment below may be useful here].
answered Jan 18 at 6:22
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
edited Jan 18 at 9:29
answered Jan 18 at 6:22
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 18 at 6:22
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
answered Jan 18 at 6:22
Kavi Rama MurthyKavi Rama Murthy
62.9k42362
62.9k42362
$begingroup$
One also needs to remark that the closure of a totally bounded set is still totally bounded.
$endgroup$
– Henno Brandsma
Jan 18 at 9:22
$begingroup$
@HennoBrandsma I am using a theorem which says that the closure of a subset $A$ of a complete metric space is compact iff $A$ is totally bounded. Yes, the proof of this uses the fact that closure of totally bounded set is totally bounded.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:28
add a comment |
$begingroup$
One also needs to remark that the closure of a totally bounded set is still totally bounded.
$endgroup$
– Henno Brandsma
Jan 18 at 9:22
$begingroup$
@HennoBrandsma I am using a theorem which says that the closure of a subset $A$ of a complete metric space is compact iff $A$ is totally bounded. Yes, the proof of this uses the fact that closure of totally bounded set is totally bounded.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:28
$begingroup$
One also needs to remark that the closure of a totally bounded set is still totally bounded.
$endgroup$
– Henno Brandsma
Jan 18 at 9:22
$begingroup$
One also needs to remark that the closure of a totally bounded set is still totally bounded.
$endgroup$
– Henno Brandsma
Jan 18 at 9:22
$begingroup$
@HennoBrandsma I am using a theorem which says that the closure of a subset $A$ of a complete metric space is compact iff $A$ is totally bounded. Yes, the proof of this uses the fact that closure of totally bounded set is totally bounded.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:28
$begingroup$
@HennoBrandsma I am using a theorem which says that the closure of a subset $A$ of a complete metric space is compact iff $A$ is totally bounded. Yes, the proof of this uses the fact that closure of totally bounded set is totally bounded.
$endgroup$
– Kavi Rama Murthy
Jan 18 at 9:28
add a comment |
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$begingroup$
I would suggest taking the intersection of all sets $A_varepsilon$. At first glance, this looks like the closure of $A$.
$endgroup$
– Lubin
Jan 18 at 6:07
$begingroup$
@Lubin why would any finite subfamily of these $A_varepsilon$ intersect?
$endgroup$
– Henno Brandsma
Jan 18 at 9:20
$begingroup$
@HennoBrandsma, it seems, after a second glance, that my suggestion is worthless. Thanks for encouraging me to think.
$endgroup$
– Lubin
Jan 18 at 14:51