Is it true that every totally bounded set in a metric space is compact?
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Every compact set is totally bounded, but can we say that every totally bounded set is compact?
I'm a beginner in metric space. My thinking is that a totally bounded set behaves like a finite set and in some sense it is small. So it is very much like a compact set.
Someone please help me to clear my doubts. Thanks.
real-analysis general-topology metric-spaces
asked Dec 8 '18 at 13:32
Arjun BanerjeeArjun Banerjee
43610
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|
show 3 more comments
$begingroup$
Every compact set is totally bounded, but can we say that every totally bounded set is compact?
I'm a beginner in metric space. My thinking is that a totally bounded set behaves like a finite set and in some sense it is small. So it is very much like a compact set.
Someone please help me to clear my doubts. Thanks.
real-analysis general-topology metric-spaces
asked Dec 8 '18 at 13:32
Arjun BanerjeeArjun Banerjee
43610
$endgroup$
1
$begingroup$
I believe the open ball is totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:37
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Oh! yes, I didn't think of that. But one more question why are we taking open balls of arbitrary diameter? Why don't we take closed balls of arbitrary diameter?
$endgroup$
– Arjun Banerjee
Dec 8 '18 at 13:41
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closed balls are compact... (In $mathbb{R}^n$ at least)
$endgroup$
– Yanko
Dec 8 '18 at 13:41
1
$begingroup$
@ArjunBanerjee You ask what happens if we change the definition of totally bounded by replacing open balls with closed balls? In this case the open ball is still totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:47
1
$begingroup$
@ArjunBanerjee It doesn't really matter, we have $B(x_0,r/2) subset B[x_0,r] subset B(x_0,2r)$ anyway. Of course, here I used $B[x_0,r]$ to denote a closed ball of radius $r$.
$endgroup$
– BigbearZzz
Dec 8 '18 at 13:48
|
show 3 more comments
$begingroup$
Every compact set is totally bounded, but can we say that every totally bounded set is compact?
I'm a beginner in metric space. My thinking is that a totally bounded set behaves like a finite set and in some sense it is small. So it is very much like a compact set.
Someone please help me to clear my doubts. Thanks.
real-analysis general-topology metric-spaces
asked Dec 8 '18 at 13:32
Arjun BanerjeeArjun Banerjee
43610
$endgroup$
Every compact set is totally bounded, but can we say that every totally bounded set is compact?
I'm a beginner in metric space. My thinking is that a totally bounded set behaves like a finite set and in some sense it is small. So it is very much like a compact set.
Someone please help me to clear my doubts. Thanks.
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked Dec 8 '18 at 13:32
Arjun BanerjeeArjun Banerjee
43610
asked Dec 8 '18 at 13:32
Arjun BanerjeeArjun Banerjee
43610
asked Dec 8 '18 at 13:32
Arjun BanerjeeArjun Banerjee
43610
asked Dec 8 '18 at 13:32
Arjun BanerjeeArjun Banerjee
43610
asked Dec 8 '18 at 13:32
Arjun BanerjeeArjun Banerjee
43610
43610
1
$begingroup$
I believe the open ball is totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:37
$begingroup$
Oh! yes, I didn't think of that. But one more question why are we taking open balls of arbitrary diameter? Why don't we take closed balls of arbitrary diameter?
$endgroup$
– Arjun Banerjee
Dec 8 '18 at 13:41
$begingroup$
closed balls are compact... (In $mathbb{R}^n$ at least)
$endgroup$
– Yanko
Dec 8 '18 at 13:41
1
$begingroup$
@ArjunBanerjee You ask what happens if we change the definition of totally bounded by replacing open balls with closed balls? In this case the open ball is still totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:47
1
$begingroup$
@ArjunBanerjee It doesn't really matter, we have $B(x_0,r/2) subset B[x_0,r] subset B(x_0,2r)$ anyway. Of course, here I used $B[x_0,r]$ to denote a closed ball of radius $r$.
$endgroup$
– BigbearZzz
Dec 8 '18 at 13:48
|
show 3 more comments
1
$begingroup$
I believe the open ball is totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:37
$begingroup$
Oh! yes, I didn't think of that. But one more question why are we taking open balls of arbitrary diameter? Why don't we take closed balls of arbitrary diameter?
$endgroup$
– Arjun Banerjee
Dec 8 '18 at 13:41
$begingroup$
closed balls are compact... (In $mathbb{R}^n$ at least)
$endgroup$
– Yanko
Dec 8 '18 at 13:41
1
$begingroup$
@ArjunBanerjee You ask what happens if we change the definition of totally bounded by replacing open balls with closed balls? In this case the open ball is still totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:47
1
$begingroup$
@ArjunBanerjee It doesn't really matter, we have $B(x_0,r/2) subset B[x_0,r] subset B(x_0,2r)$ anyway. Of course, here I used $B[x_0,r]$ to denote a closed ball of radius $r$.
$endgroup$
– BigbearZzz
Dec 8 '18 at 13:48
1
1
$begingroup$
I believe the open ball is totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:37
$begingroup$
I believe the open ball is totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:37
$begingroup$
Oh! yes, I didn't think of that. But one more question why are we taking open balls of arbitrary diameter? Why don't we take closed balls of arbitrary diameter?
$endgroup$
– Arjun Banerjee
Dec 8 '18 at 13:41
$begingroup$
Oh! yes, I didn't think of that. But one more question why are we taking open balls of arbitrary diameter? Why don't we take closed balls of arbitrary diameter?
$endgroup$
– Arjun Banerjee
Dec 8 '18 at 13:41
$begingroup$
closed balls are compact... (In $mathbb{R}^n$ at least)
$endgroup$
– Yanko
Dec 8 '18 at 13:41
$begingroup$
closed balls are compact... (In $mathbb{R}^n$ at least)
$endgroup$
– Yanko
Dec 8 '18 at 13:41
1
1
$begingroup$
@ArjunBanerjee You ask what happens if we change the definition of totally bounded by replacing open balls with closed balls? In this case the open ball is still totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:47
$begingroup$
@ArjunBanerjee You ask what happens if we change the definition of totally bounded by replacing open balls with closed balls? In this case the open ball is still totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:47
1
1
$begingroup$
@ArjunBanerjee It doesn't really matter, we have $B(x_0,r/2) subset B[x_0,r] subset B(x_0,2r)$ anyway. Of course, here I used $B[x_0,r]$ to denote a closed ball of radius $r$.
$endgroup$
– BigbearZzz
Dec 8 '18 at 13:48
$begingroup$
@ArjunBanerjee It doesn't really matter, we have $B(x_0,r/2) subset B[x_0,r] subset B(x_0,2r)$ anyway. Of course, here I used $B[x_0,r]$ to denote a closed ball of radius $r$.
$endgroup$
– BigbearZzz
Dec 8 '18 at 13:48
|
show 3 more comments
1 Answer
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No, that is not enough (but almost).
Consider the set $(0,1)$ in the metric space $Bbb R$, it is totally bounded but not compact. However, it is well known that totally boundedness & completeness is equivalent to compactness. You can read more about that here.
answered Dec 8 '18 at 13:38
BigbearZzzBigbearZzz
8,51721652
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$begingroup$
No, that is not enough (but almost).
Consider the set $(0,1)$ in the metric space $Bbb R$, it is totally bounded but not compact. However, it is well known that totally boundedness & completeness is equivalent to compactness. You can read more about that here.
answered Dec 8 '18 at 13:38
BigbearZzzBigbearZzz
8,51721652
$endgroup$
add a comment |
$begingroup$
No, that is not enough (but almost).
Consider the set $(0,1)$ in the metric space $Bbb R$, it is totally bounded but not compact. However, it is well known that totally boundedness & completeness is equivalent to compactness. You can read more about that here.
answered Dec 8 '18 at 13:38
BigbearZzzBigbearZzz
8,51721652
$endgroup$
add a comment |
$begingroup$
No, that is not enough (but almost).
Consider the set $(0,1)$ in the metric space $Bbb R$, it is totally bounded but not compact. However, it is well known that totally boundedness & completeness is equivalent to compactness. You can read more about that here.
answered Dec 8 '18 at 13:38
BigbearZzzBigbearZzz
8,51721652
$endgroup$
No, that is not enough (but almost).
Consider the set $(0,1)$ in the metric space $Bbb R$, it is totally bounded but not compact. However, it is well known that totally boundedness & completeness is equivalent to compactness. You can read more about that here.
answered Dec 8 '18 at 13:38
BigbearZzzBigbearZzz
8,51721652
edited Dec 14 '18 at 8:48
answered Dec 8 '18 at 13:38
BigbearZzzBigbearZzz
8,51721652
answered Dec 8 '18 at 13:38
BigbearZzzBigbearZzz
8,51721652
answered Dec 8 '18 at 13:38
BigbearZzzBigbearZzz
8,51721652
8,51721652
add a comment |
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1
$begingroup$
I believe the open ball is totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:37
$begingroup$
Oh! yes, I didn't think of that. But one more question why are we taking open balls of arbitrary diameter? Why don't we take closed balls of arbitrary diameter?
$endgroup$
– Arjun Banerjee
Dec 8 '18 at 13:41
$begingroup$
closed balls are compact... (In $mathbb{R}^n$ at least)
$endgroup$
– Yanko
Dec 8 '18 at 13:41
1
$begingroup$
@ArjunBanerjee You ask what happens if we change the definition of totally bounded by replacing open balls with closed balls? In this case the open ball is still totally bounded but it is not compact.
$endgroup$
– Yanko
Dec 8 '18 at 13:47
1
$begingroup$
@ArjunBanerjee It doesn't really matter, we have $B(x_0,r/2) subset B[x_0,r] subset B(x_0,2r)$ anyway. Of course, here I used $B[x_0,r]$ to denote a closed ball of radius $r$.
$endgroup$
– BigbearZzz
Dec 8 '18 at 13:48