Are these implications regarding compactness correct?
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Is the following true and if not how is it incorrect :
a subset of a metric space being compact $LeftarrowRightarrow$ it is complete and totally bounded $Rightarrow$ it is closed and totally bounded.
general-topology metric-spaces
edited Aug 7 at 17:43
Andrés E. Caicedo
63.2k7151236
asked Aug 7 at 17:09
exodius
772315
add a comment |Â
up vote
0
down vote
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Is the following true and if not how is it incorrect :
a subset of a metric space being compact $LeftarrowRightarrow$ it is complete and totally bounded $Rightarrow$ it is closed and totally bounded.
general-topology metric-spaces
edited Aug 7 at 17:43
Andrés E. Caicedo
63.2k7151236
asked Aug 7 at 17:09
exodius
772315
Everything is correct. Note that a metric space for itself is always closed.
– amsmath
Aug 7 at 17:16
It is quite unclear what you mean by "closed" when you talk about a "space".
– Saucy O'Path
Aug 7 at 17:18
2
First, on the left I presume you mean "a metric space being compact"; compactness is a property of more general topological spaces, and these implications do not make sense for a topological space which is not a metric space. Second, being "closed" is not a property of a metric space, it is instead a property of a subspace, so your second arrow does not make sense.
– Lee Mosher
Aug 7 at 17:18
@LeeMosher ah okay so if I change space to subset of a metric space then everything is okay ?
– exodius
Aug 7 at 17:19
Yes, everything will be okay.
– Lee Mosher
Aug 7 at 17:21
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is the following true and if not how is it incorrect :
a subset of a metric space being compact $LeftarrowRightarrow$ it is complete and totally bounded $Rightarrow$ it is closed and totally bounded.
general-topology metric-spaces
edited Aug 7 at 17:43
Andrés E. Caicedo
63.2k7151236
asked Aug 7 at 17:09
exodius
772315
Is the following true and if not how is it incorrect :
a subset of a metric space being compact $LeftarrowRightarrow$ it is complete and totally bounded $Rightarrow$ it is closed and totally bounded.
general-topology metric-spaces
edited Aug 7 at 17:43
Andrés E. Caicedo
63.2k7151236
asked Aug 7 at 17:09
exodius
772315
edited Aug 7 at 17:43
Andrés E. Caicedo
63.2k7151236
edited Aug 7 at 17:43
Andrés E. Caicedo
63.2k7151236
edited Aug 7 at 17:43
Andrés E. Caicedo
63.2k7151236
63.2k7151236
asked Aug 7 at 17:09
exodius
772315
asked Aug 7 at 17:09
exodius
772315
asked Aug 7 at 17:09
exodius
772315
772315
Everything is correct. Note that a metric space for itself is always closed.
– amsmath
Aug 7 at 17:16
It is quite unclear what you mean by "closed" when you talk about a "space".
– Saucy O'Path
Aug 7 at 17:18
2
First, on the left I presume you mean "a metric space being compact"; compactness is a property of more general topological spaces, and these implications do not make sense for a topological space which is not a metric space. Second, being "closed" is not a property of a metric space, it is instead a property of a subspace, so your second arrow does not make sense.
– Lee Mosher
Aug 7 at 17:18
@LeeMosher ah okay so if I change space to subset of a metric space then everything is okay ?
– exodius
Aug 7 at 17:19
Yes, everything will be okay.
– Lee Mosher
Aug 7 at 17:21
add a comment |Â
Everything is correct. Note that a metric space for itself is always closed.
– amsmath
Aug 7 at 17:16
It is quite unclear what you mean by "closed" when you talk about a "space".
– Saucy O'Path
Aug 7 at 17:18
2
First, on the left I presume you mean "a metric space being compact"; compactness is a property of more general topological spaces, and these implications do not make sense for a topological space which is not a metric space. Second, being "closed" is not a property of a metric space, it is instead a property of a subspace, so your second arrow does not make sense.
– Lee Mosher
Aug 7 at 17:18
@LeeMosher ah okay so if I change space to subset of a metric space then everything is okay ?
– exodius
Aug 7 at 17:19
Yes, everything will be okay.
– Lee Mosher
Aug 7 at 17:21
Everything is correct. Note that a metric space for itself is always closed.
– amsmath
Aug 7 at 17:16
Everything is correct. Note that a metric space for itself is always closed.
– amsmath
Aug 7 at 17:16
It is quite unclear what you mean by "closed" when you talk about a "space".
– Saucy O'Path
Aug 7 at 17:18
It is quite unclear what you mean by "closed" when you talk about a "space".
– Saucy O'Path
Aug 7 at 17:18
2
2
First, on the left I presume you mean "a metric space being compact"; compactness is a property of more general topological spaces, and these implications do not make sense for a topological space which is not a metric space. Second, being "closed" is not a property of a metric space, it is instead a property of a subspace, so your second arrow does not make sense.
– Lee Mosher
Aug 7 at 17:18
First, on the left I presume you mean "a metric space being compact"; compactness is a property of more general topological spaces, and these implications do not make sense for a topological space which is not a metric space. Second, being "closed" is not a property of a metric space, it is instead a property of a subspace, so your second arrow does not make sense.
– Lee Mosher
Aug 7 at 17:18
@LeeMosher ah okay so if I change space to subset of a metric space then everything is okay ?
– exodius
Aug 7 at 17:19
@LeeMosher ah okay so if I change space to subset of a metric space then everything is okay ?
– exodius
Aug 7 at 17:19
Yes, everything will be okay.
– Lee Mosher
Aug 7 at 17:21
Yes, everything will be okay.
– Lee Mosher
Aug 7 at 17:21
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your statement is essentially correct, but can be stated more precisely:
Let $(X,d)$ be a metric space, and let $A subset X$ be a metric subspace, so we give $A$ the metric $d_A$ induced from $d$ (just the restriction of $d$ to $A times A$).
Then indeed $(A, d_A)$ is compact iff it is complete and totally bounded.
The compactness (or completeness plus total boundedness) does imply that $A$ is closed in $(X,d)$ (and also still totally bounded), but $A$ being closed in $(X,d)$ and totally bounded (under $d_A$) does not imply compactness of $(A,d_A)$. This would be true if $(X,d)$ were complete to start with, but not otherwise, e.g. take $X=(0,1)$ in the Euclidean metric and $A=X$, which is closed in $X$ (trivially) and totally bounded, but not compact.
answered Aug 8 at 9:59
Henno Brandsma
91.7k342100
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your statement is essentially correct, but can be stated more precisely:
Let $(X,d)$ be a metric space, and let $A subset X$ be a metric subspace, so we give $A$ the metric $d_A$ induced from $d$ (just the restriction of $d$ to $A times A$).
Then indeed $(A, d_A)$ is compact iff it is complete and totally bounded.
The compactness (or completeness plus total boundedness) does imply that $A$ is closed in $(X,d)$ (and also still totally bounded), but $A$ being closed in $(X,d)$ and totally bounded (under $d_A$) does not imply compactness of $(A,d_A)$. This would be true if $(X,d)$ were complete to start with, but not otherwise, e.g. take $X=(0,1)$ in the Euclidean metric and $A=X$, which is closed in $X$ (trivially) and totally bounded, but not compact.
answered Aug 8 at 9:59
Henno Brandsma
91.7k342100
add a comment |Â
up vote
1
down vote
accepted
Your statement is essentially correct, but can be stated more precisely:
Let $(X,d)$ be a metric space, and let $A subset X$ be a metric subspace, so we give $A$ the metric $d_A$ induced from $d$ (just the restriction of $d$ to $A times A$).
Then indeed $(A, d_A)$ is compact iff it is complete and totally bounded.
The compactness (or completeness plus total boundedness) does imply that $A$ is closed in $(X,d)$ (and also still totally bounded), but $A$ being closed in $(X,d)$ and totally bounded (under $d_A$) does not imply compactness of $(A,d_A)$. This would be true if $(X,d)$ were complete to start with, but not otherwise, e.g. take $X=(0,1)$ in the Euclidean metric and $A=X$, which is closed in $X$ (trivially) and totally bounded, but not compact.
answered Aug 8 at 9:59
Henno Brandsma
91.7k342100
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your statement is essentially correct, but can be stated more precisely:
Let $(X,d)$ be a metric space, and let $A subset X$ be a metric subspace, so we give $A$ the metric $d_A$ induced from $d$ (just the restriction of $d$ to $A times A$).
Then indeed $(A, d_A)$ is compact iff it is complete and totally bounded.
The compactness (or completeness plus total boundedness) does imply that $A$ is closed in $(X,d)$ (and also still totally bounded), but $A$ being closed in $(X,d)$ and totally bounded (under $d_A$) does not imply compactness of $(A,d_A)$. This would be true if $(X,d)$ were complete to start with, but not otherwise, e.g. take $X=(0,1)$ in the Euclidean metric and $A=X$, which is closed in $X$ (trivially) and totally bounded, but not compact.
answered Aug 8 at 9:59
Henno Brandsma
91.7k342100
Your statement is essentially correct, but can be stated more precisely:
Let $(X,d)$ be a metric space, and let $A subset X$ be a metric subspace, so we give $A$ the metric $d_A$ induced from $d$ (just the restriction of $d$ to $A times A$).
Then indeed $(A, d_A)$ is compact iff it is complete and totally bounded.
The compactness (or completeness plus total boundedness) does imply that $A$ is closed in $(X,d)$ (and also still totally bounded), but $A$ being closed in $(X,d)$ and totally bounded (under $d_A$) does not imply compactness of $(A,d_A)$. This would be true if $(X,d)$ were complete to start with, but not otherwise, e.g. take $X=(0,1)$ in the Euclidean metric and $A=X$, which is closed in $X$ (trivially) and totally bounded, but not compact.
answered Aug 8 at 9:59
Henno Brandsma
91.7k342100
answered Aug 8 at 9:59
Henno Brandsma
91.7k342100
answered Aug 8 at 9:59
Henno Brandsma
91.7k342100
answered Aug 8 at 9:59
Henno Brandsma
91.7k342100
91.7k342100
add a comment |Â
add a comment |Â
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Everything is correct. Note that a metric space for itself is always closed.
– amsmath
Aug 7 at 17:16
It is quite unclear what you mean by "closed" when you talk about a "space".
– Saucy O'Path
Aug 7 at 17:18
2
First, on the left I presume you mean "a metric space being compact"; compactness is a property of more general topological spaces, and these implications do not make sense for a topological space which is not a metric space. Second, being "closed" is not a property of a metric space, it is instead a property of a subspace, so your second arrow does not make sense.
– Lee Mosher
Aug 7 at 17:18
@LeeMosher ah okay so if I change space to subset of a metric space then everything is okay ?
– exodius
Aug 7 at 17:19
Yes, everything will be okay.
– Lee Mosher
Aug 7 at 17:21