For a subset $A$ of a metric space which of the following implies the other three ?
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0
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For a subset $A$ of a metric space which of the following implies the other three ?
$a)$ $A$ is closed
$b)$ $A$ is bounded
$c)$Closure of $B$ is compact ,for every $B subseteq A$
$d)$ $A$ is compact
I thinks option d) will correct .
is it true??
general-topology
asked Nov 23 at 13:44
Messi fifa
51611
add a comment |
up vote
0
down vote
favorite
For a subset $A$ of a metric space which of the following implies the other three ?
$a)$ $A$ is closed
$b)$ $A$ is bounded
$c)$Closure of $B$ is compact ,for every $B subseteq A$
$d)$ $A$ is compact
I thinks option d) will correct .
is it true??
general-topology
asked Nov 23 at 13:44
Messi fifa
51611
1
Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
– John_Wick
Nov 23 at 13:50
@John_Wick thanks u
– Messi fifa
Nov 23 at 13:51
Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
– John Nash
Nov 23 at 13:54
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For a subset $A$ of a metric space which of the following implies the other three ?
$a)$ $A$ is closed
$b)$ $A$ is bounded
$c)$Closure of $B$ is compact ,for every $B subseteq A$
$d)$ $A$ is compact
I thinks option d) will correct .
is it true??
general-topology
asked Nov 23 at 13:44
Messi fifa
51611
For a subset $A$ of a metric space which of the following implies the other three ?
$a)$ $A$ is closed
$b)$ $A$ is bounded
$c)$Closure of $B$ is compact ,for every $B subseteq A$
$d)$ $A$ is compact
I thinks option d) will correct .
is it true??
general-topology
general-topology
asked Nov 23 at 13:44
Messi fifa
51611
asked Nov 23 at 13:44
Messi fifa
51611
asked Nov 23 at 13:44
Messi fifa
51611
asked Nov 23 at 13:44
Messi fifa
51611
asked Nov 23 at 13:44
Messi fifa
51611
51611
1
Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
– John_Wick
Nov 23 at 13:50
@John_Wick thanks u
– Messi fifa
Nov 23 at 13:51
Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
– John Nash
Nov 23 at 13:54
add a comment |
1
Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
– John_Wick
Nov 23 at 13:50
@John_Wick thanks u
– Messi fifa
Nov 23 at 13:51
Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
– John Nash
Nov 23 at 13:54
1
1
Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
– John_Wick
Nov 23 at 13:50
Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
– John_Wick
Nov 23 at 13:50
@John_Wick thanks u
– Messi fifa
Nov 23 at 13:51
@John_Wick thanks u
– Messi fifa
Nov 23 at 13:51
Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
– John Nash
Nov 23 at 13:54
Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
– John Nash
Nov 23 at 13:54
add a comment |
1 Answer
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(d) is right, as a compact set is closed and bounded in any metric space.
and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.
answered Nov 23 at 17:07
Henno Brandsma
103k345112
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
(d) is right, as a compact set is closed and bounded in any metric space.
and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.
answered Nov 23 at 17:07
Henno Brandsma
103k345112
add a comment |
up vote
2
down vote
accepted
(d) is right, as a compact set is closed and bounded in any metric space.
and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.
answered Nov 23 at 17:07
Henno Brandsma
103k345112
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
(d) is right, as a compact set is closed and bounded in any metric space.
and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.
answered Nov 23 at 17:07
Henno Brandsma
103k345112
(d) is right, as a compact set is closed and bounded in any metric space.
and also if $A$ is compact then for $B subseteq A$, $overline{B}$ is closed in $overline{A}=A$ and hence compact too.
answered Nov 23 at 17:07
Henno Brandsma
103k345112
answered Nov 23 at 17:07
Henno Brandsma
103k345112
answered Nov 23 at 17:07
Henno Brandsma
103k345112
answered Nov 23 at 17:07
Henno Brandsma
103k345112
103k345112
add a comment |
add a comment |
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Yes, it's true. (d) is the strongest condition you can have among them, (d) implies all the other.
– John_Wick
Nov 23 at 13:50
@John_Wick thanks u
– Messi fifa
Nov 23 at 13:51
Obviously If $A$ is compact then it implies $A$ is closed and bounded then there is a subset B whose closure is compact in fact any closed closed subset of a compact set in a (metric) space X is compact.
– John Nash
Nov 23 at 13:54