Let's consider the metric space $BbbQ$ with the euclidian metric of $BbbR$
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Understanding the theorem:
Let's consider the metric space $BbbQ$ with the euclidian metric from $BbbR$.
Let $a,bin$ $BbbRsetminusBbbQ$ with $a<b$ and $S=(a,b)capBbbQ$ . Prove that $S$ is closed and bounded in $BbbQ$, but not compact.
My question is why $S$ is not compact if Heine-Borel's theorem tells me that if a set is closed and bounded then the set is compact.
general-topology
edited Sep 8 at 11:29
Javi
2,2731725
asked Sep 8 at 4:25
camilo
266
 |Â
show 3 more comments
up vote
1
down vote
favorite
Understanding the theorem:
Let's consider the metric space $BbbQ$ with the euclidian metric from $BbbR$.
Let $a,bin$ $BbbRsetminusBbbQ$ with $a<b$ and $S=(a,b)capBbbQ$ . Prove that $S$ is closed and bounded in $BbbQ$, but not compact.
My question is why $S$ is not compact if Heine-Borel's theorem tells me that if a set is closed and bounded then the set is compact.
general-topology
edited Sep 8 at 11:29
Javi
2,2731725
asked Sep 8 at 4:25
camilo
266
Heine-Borel's theorem tells me that if a set is closed and bounded then the set is connected - you mean the set is compact.
– Brevan Ellefsen
Sep 8 at 4:28
The Heine-Borel theorem is about the topology on $R^n.$ However, note you are proving $S$ is compact in $mathbbQ$, not in $mathbbR$!
– Brevan Ellefsen
Sep 8 at 4:30
Please fix the typo. connected -> compact. Note that the Heine Borel theorem applies to $mathbbR^n$. Note that $S$ is not complete, so there are Cauchy sequences that do not have a limit.
– copper.hat
Sep 8 at 4:37
Yes! thank you, I thought the same thing but I dont was not sure.
– camilo
Sep 8 at 4:38
1
@dmtri $a$ must be irrational, and so is $b$. Hence, $S$ may be written as $[a,b]cap mathbb Q$, which is closed in $mathbb Q$ for the induced topology.
– Suzet
Sep 8 at 5:20
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Understanding the theorem:
Let's consider the metric space $BbbQ$ with the euclidian metric from $BbbR$.
Let $a,bin$ $BbbRsetminusBbbQ$ with $a<b$ and $S=(a,b)capBbbQ$ . Prove that $S$ is closed and bounded in $BbbQ$, but not compact.
My question is why $S$ is not compact if Heine-Borel's theorem tells me that if a set is closed and bounded then the set is compact.
general-topology
edited Sep 8 at 11:29
Javi
2,2731725
asked Sep 8 at 4:25
camilo
266
Understanding the theorem:
Let's consider the metric space $BbbQ$ with the euclidian metric from $BbbR$.
Let $a,bin$ $BbbRsetminusBbbQ$ with $a<b$ and $S=(a,b)capBbbQ$ . Prove that $S$ is closed and bounded in $BbbQ$, but not compact.
My question is why $S$ is not compact if Heine-Borel's theorem tells me that if a set is closed and bounded then the set is compact.
general-topology
general-topology
edited Sep 8 at 11:29
Javi
2,2731725
asked Sep 8 at 4:25
camilo
266
edited Sep 8 at 11:29
Javi
2,2731725
asked Sep 8 at 4:25
camilo
266
edited Sep 8 at 11:29
Javi
2,2731725
edited Sep 8 at 11:29
Javi
2,2731725
edited Sep 8 at 11:29
Javi
2,2731725
2,2731725
asked Sep 8 at 4:25
camilo
266
asked Sep 8 at 4:25
camilo
266
asked Sep 8 at 4:25
camilo
266
266
Heine-Borel's theorem tells me that if a set is closed and bounded then the set is connected - you mean the set is compact.
– Brevan Ellefsen
Sep 8 at 4:28
The Heine-Borel theorem is about the topology on $R^n.$ However, note you are proving $S$ is compact in $mathbbQ$, not in $mathbbR$!
– Brevan Ellefsen
Sep 8 at 4:30
Please fix the typo. connected -> compact. Note that the Heine Borel theorem applies to $mathbbR^n$. Note that $S$ is not complete, so there are Cauchy sequences that do not have a limit.
– copper.hat
Sep 8 at 4:37
Yes! thank you, I thought the same thing but I dont was not sure.
– camilo
Sep 8 at 4:38
1
@dmtri $a$ must be irrational, and so is $b$. Hence, $S$ may be written as $[a,b]cap mathbb Q$, which is closed in $mathbb Q$ for the induced topology.
– Suzet
Sep 8 at 5:20
 |Â
show 3 more comments
Heine-Borel's theorem tells me that if a set is closed and bounded then the set is connected - you mean the set is compact.
– Brevan Ellefsen
Sep 8 at 4:28
The Heine-Borel theorem is about the topology on $R^n.$ However, note you are proving $S$ is compact in $mathbbQ$, not in $mathbbR$!
– Brevan Ellefsen
Sep 8 at 4:30
Please fix the typo. connected -> compact. Note that the Heine Borel theorem applies to $mathbbR^n$. Note that $S$ is not complete, so there are Cauchy sequences that do not have a limit.
– copper.hat
Sep 8 at 4:37
Yes! thank you, I thought the same thing but I dont was not sure.
– camilo
Sep 8 at 4:38
1
@dmtri $a$ must be irrational, and so is $b$. Hence, $S$ may be written as $[a,b]cap mathbb Q$, which is closed in $mathbb Q$ for the induced topology.
– Suzet
Sep 8 at 5:20
Heine-Borel's theorem tells me that if a set is closed and bounded then the set is connected - you mean the set is compact.
– Brevan Ellefsen
Sep 8 at 4:28
Heine-Borel's theorem tells me that if a set is closed and bounded then the set is connected - you mean the set is compact.
– Brevan Ellefsen
Sep 8 at 4:28
The Heine-Borel theorem is about the topology on $R^n.$ However, note you are proving $S$ is compact in $mathbbQ$, not in $mathbbR$!
– Brevan Ellefsen
Sep 8 at 4:30
The Heine-Borel theorem is about the topology on $R^n.$ However, note you are proving $S$ is compact in $mathbbQ$, not in $mathbbR$!
– Brevan Ellefsen
Sep 8 at 4:30
Please fix the typo. connected -> compact. Note that the Heine Borel theorem applies to $mathbbR^n$. Note that $S$ is not complete, so there are Cauchy sequences that do not have a limit.
– copper.hat
Sep 8 at 4:37
Please fix the typo. connected -> compact. Note that the Heine Borel theorem applies to $mathbbR^n$. Note that $S$ is not complete, so there are Cauchy sequences that do not have a limit.
– copper.hat
Sep 8 at 4:37
Yes! thank you, I thought the same thing but I dont was not sure.
– camilo
Sep 8 at 4:38
Yes! thank you, I thought the same thing but I dont was not sure.
– camilo
Sep 8 at 4:38
1
1
@dmtri $a$ must be irrational, and so is $b$. Hence, $S$ may be written as $[a,b]cap mathbb Q$, which is closed in $mathbb Q$ for the induced topology.
– Suzet
Sep 8 at 5:20
@dmtri $a$ must be irrational, and so is $b$. Hence, $S$ may be written as $[a,b]cap mathbb Q$, which is closed in $mathbb Q$ for the induced topology.
– Suzet
Sep 8 at 5:20
 |Â
show 3 more comments
1 Answer
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Heine-Borel's theorem talks about subsets of $mathbbR^n$ but you are dealing with a subset of $mathbbQ$. While being bounded doesn't depend on the choice of the "big" space being closed does. For example $(-sqrt2, sqrt2)capmathbbQ$ is closed in $mathbbQ$ but not in $mathbbR$.
So back to your question. For any real number $rinmathbbR$ there exists a sequence $(q_n)subseteqmathbbQ$ convergent to $r$. Also if $a,binmathbbR$, $a<b$ then there exists a irrational $zinmathbbRbackslashmathbbQ$ such that $a<z<b$. These two facts together imply $(a,b)capmathbbQ$ cannot be compact for any $a<b$.
Now your $S$ is obviously bounded. It is also closed because $(a,b)capmathbbQ=[a,b]capmathbbQ$ if $a,b$ are irrational.
answered Sep 8 at 7:19
freakish
8,8831524
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Heine-Borel's theorem talks about subsets of $mathbbR^n$ but you are dealing with a subset of $mathbbQ$. While being bounded doesn't depend on the choice of the "big" space being closed does. For example $(-sqrt2, sqrt2)capmathbbQ$ is closed in $mathbbQ$ but not in $mathbbR$.
So back to your question. For any real number $rinmathbbR$ there exists a sequence $(q_n)subseteqmathbbQ$ convergent to $r$. Also if $a,binmathbbR$, $a<b$ then there exists a irrational $zinmathbbRbackslashmathbbQ$ such that $a<z<b$. These two facts together imply $(a,b)capmathbbQ$ cannot be compact for any $a<b$.
Now your $S$ is obviously bounded. It is also closed because $(a,b)capmathbbQ=[a,b]capmathbbQ$ if $a,b$ are irrational.
answered Sep 8 at 7:19
freakish
8,8831524
add a comment |Â
up vote
3
down vote
Heine-Borel's theorem talks about subsets of $mathbbR^n$ but you are dealing with a subset of $mathbbQ$. While being bounded doesn't depend on the choice of the "big" space being closed does. For example $(-sqrt2, sqrt2)capmathbbQ$ is closed in $mathbbQ$ but not in $mathbbR$.
So back to your question. For any real number $rinmathbbR$ there exists a sequence $(q_n)subseteqmathbbQ$ convergent to $r$. Also if $a,binmathbbR$, $a<b$ then there exists a irrational $zinmathbbRbackslashmathbbQ$ such that $a<z<b$. These two facts together imply $(a,b)capmathbbQ$ cannot be compact for any $a<b$.
Now your $S$ is obviously bounded. It is also closed because $(a,b)capmathbbQ=[a,b]capmathbbQ$ if $a,b$ are irrational.
answered Sep 8 at 7:19
freakish
8,8831524
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Heine-Borel's theorem talks about subsets of $mathbbR^n$ but you are dealing with a subset of $mathbbQ$. While being bounded doesn't depend on the choice of the "big" space being closed does. For example $(-sqrt2, sqrt2)capmathbbQ$ is closed in $mathbbQ$ but not in $mathbbR$.
So back to your question. For any real number $rinmathbbR$ there exists a sequence $(q_n)subseteqmathbbQ$ convergent to $r$. Also if $a,binmathbbR$, $a<b$ then there exists a irrational $zinmathbbRbackslashmathbbQ$ such that $a<z<b$. These two facts together imply $(a,b)capmathbbQ$ cannot be compact for any $a<b$.
Now your $S$ is obviously bounded. It is also closed because $(a,b)capmathbbQ=[a,b]capmathbbQ$ if $a,b$ are irrational.
answered Sep 8 at 7:19
freakish
8,8831524
Heine-Borel's theorem talks about subsets of $mathbbR^n$ but you are dealing with a subset of $mathbbQ$. While being bounded doesn't depend on the choice of the "big" space being closed does. For example $(-sqrt2, sqrt2)capmathbbQ$ is closed in $mathbbQ$ but not in $mathbbR$.
So back to your question. For any real number $rinmathbbR$ there exists a sequence $(q_n)subseteqmathbbQ$ convergent to $r$. Also if $a,binmathbbR$, $a<b$ then there exists a irrational $zinmathbbRbackslashmathbbQ$ such that $a<z<b$. These two facts together imply $(a,b)capmathbbQ$ cannot be compact for any $a<b$.
Now your $S$ is obviously bounded. It is also closed because $(a,b)capmathbbQ=[a,b]capmathbbQ$ if $a,b$ are irrational.
answered Sep 8 at 7:19
freakish
8,8831524
answered Sep 8 at 7:19
freakish
8,8831524
answered Sep 8 at 7:19
freakish
8,8831524
answered Sep 8 at 7:19
freakish
8,8831524
8,8831524
add a comment |Â
add a comment |Â
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Heine-Borel's theorem tells me that if a set is closed and bounded then the set is connected - you mean the set is compact.
– Brevan Ellefsen
Sep 8 at 4:28
The Heine-Borel theorem is about the topology on $R^n.$ However, note you are proving $S$ is compact in $mathbbQ$, not in $mathbbR$!
– Brevan Ellefsen
Sep 8 at 4:30
Please fix the typo. connected -> compact. Note that the Heine Borel theorem applies to $mathbbR^n$. Note that $S$ is not complete, so there are Cauchy sequences that do not have a limit.
– copper.hat
Sep 8 at 4:37
Yes! thank you, I thought the same thing but I dont was not sure.
– camilo
Sep 8 at 4:38
1
@dmtri $a$ must be irrational, and so is $b$. Hence, $S$ may be written as $[a,b]cap mathbb Q$, which is closed in $mathbb Q$ for the induced topology.
– Suzet
Sep 8 at 5:20