Compact subet of $Bbb R^2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Subset of Cantor set that isn't compactEquality of measures, symmetric intervals, compact, bounded, measurable setsFor a compact set $Ksubset Bbb R^n $ prove the following :Is $[0,1] cap Bbb Q$ a compact subset of $Bbb Q$?Prove that a metrizable space is countably compact iff it is compact.Methods for proving if a set is open, closed or compactConnected sets in $Bbb R$ and open subsets in $Bbb R$Prove that if $mathfrak A$ contains a bounded set then $exists M in Bbb Z^+$ s.t $A_1 cap cdots A_M=∅$there is an $R >0$ such that $d(x , C) < R$ implies $x in O$. Hint: use the Heine-Borel theorem.Proof of a compact set without using Heine-Borel
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Compact subet of $Bbb R^2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Subset of Cantor set that isn't compactEquality of measures, symmetric intervals, compact, bounded, measurable setsFor a compact set $Ksubset Bbb R^n $ prove the following :Is $[0,1] cap Bbb Q$ a compact subset of $Bbb Q$?Prove that a metrizable space is countably compact iff it is compact.Methods for proving if a set is open, closed or compactConnected sets in $Bbb R$ and open subsets in $Bbb R$Prove that if $mathfrak A$ contains a bounded set then $exists M in Bbb Z^+$ s.t $A_1 cap cdots A_M=∅$there is an $R >0$ such that $d(x , C) < R$ implies $x in O$. Hint: use the Heine-Borel theorem.Proof of a compact set without using Heine-Borel
$begingroup$
If $K$ is a compact subset of $Bbb R^2$ then prove that $Ksubset [a,b]times [c,d]$ for some pair of compact intervals $[a,b]$ and $[c,d]$.
How can I prove this? Any hint?
We know that any compact set of $Bbb R$ is of the form $[a,b]$ or any finite set. But how can I figure out the subset of $Bbb R^2$ ?
real-analysis general-topology analysis metric-spaces
asked Mar 27 at 3:06
TopoTopo
330314
$endgroup$
add a comment |
$begingroup$
If $K$ is a compact subset of $Bbb R^2$ then prove that $Ksubset [a,b]times [c,d]$ for some pair of compact intervals $[a,b]$ and $[c,d]$.
How can I prove this? Any hint?
We know that any compact set of $Bbb R$ is of the form $[a,b]$ or any finite set. But how can I figure out the subset of $Bbb R^2$ ?
real-analysis general-topology analysis metric-spaces
asked Mar 27 at 3:06
TopoTopo
330314
$endgroup$
1
$begingroup$
Hint: Show $K$ must be bounded.
$endgroup$
– Theo Bendit
Mar 27 at 3:11
1
$begingroup$
There are exotic compact sets in the real line, not only intervals...
$endgroup$
– Eduardo Longa
Mar 27 at 3:13
4
$begingroup$
You have the wrong typo in the title. The correct typo for subset is sunset.
$endgroup$
– DanielWainfleet
Mar 27 at 4:48
add a comment |
$begingroup$
If $K$ is a compact subset of $Bbb R^2$ then prove that $Ksubset [a,b]times [c,d]$ for some pair of compact intervals $[a,b]$ and $[c,d]$.
How can I prove this? Any hint?
We know that any compact set of $Bbb R$ is of the form $[a,b]$ or any finite set. But how can I figure out the subset of $Bbb R^2$ ?
real-analysis general-topology analysis metric-spaces
asked Mar 27 at 3:06
TopoTopo
330314
$endgroup$
If $K$ is a compact subset of $Bbb R^2$ then prove that $Ksubset [a,b]times [c,d]$ for some pair of compact intervals $[a,b]$ and $[c,d]$.
How can I prove this? Any hint?
We know that any compact set of $Bbb R$ is of the form $[a,b]$ or any finite set. But how can I figure out the subset of $Bbb R^2$ ?
real-analysis general-topology analysis metric-spaces
real-analysis general-topology analysis metric-spaces
asked Mar 27 at 3:06
TopoTopo
330314
asked Mar 27 at 3:06
TopoTopo
330314
edited Mar 27 at 3:17
Topo
asked Mar 27 at 3:06
TopoTopo
330314
asked Mar 27 at 3:06
TopoTopo
330314
asked Mar 27 at 3:06
TopoTopo
330314
330314
1
$begingroup$
Hint: Show $K$ must be bounded.
$endgroup$
– Theo Bendit
Mar 27 at 3:11
1
$begingroup$
There are exotic compact sets in the real line, not only intervals...
$endgroup$
– Eduardo Longa
Mar 27 at 3:13
4
$begingroup$
You have the wrong typo in the title. The correct typo for subset is sunset.
$endgroup$
– DanielWainfleet
Mar 27 at 4:48
add a comment |
1
$begingroup$
Hint: Show $K$ must be bounded.
$endgroup$
– Theo Bendit
Mar 27 at 3:11
1
$begingroup$
There are exotic compact sets in the real line, not only intervals...
$endgroup$
– Eduardo Longa
Mar 27 at 3:13
4
$begingroup$
You have the wrong typo in the title. The correct typo for subset is sunset.
$endgroup$
– DanielWainfleet
Mar 27 at 4:48
1
1
$begingroup$
Hint: Show $K$ must be bounded.
$endgroup$
– Theo Bendit
Mar 27 at 3:11
$begingroup$
Hint: Show $K$ must be bounded.
$endgroup$
– Theo Bendit
Mar 27 at 3:11
1
1
$begingroup$
There are exotic compact sets in the real line, not only intervals...
$endgroup$
– Eduardo Longa
Mar 27 at 3:13
$begingroup$
There are exotic compact sets in the real line, not only intervals...
$endgroup$
– Eduardo Longa
Mar 27 at 3:13
4
4
$begingroup$
You have the wrong typo in the title. The correct typo for subset is sunset.
$endgroup$
– DanielWainfleet
Mar 27 at 4:48
$begingroup$
You have the wrong typo in the title. The correct typo for subset is sunset.
$endgroup$
– DanielWainfleet
Mar 27 at 4:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the family $left G_n right_n geqslant 1$ where each $G_n$ is the open subset of $mathbbR^2$ given by
$$
G_n := (-n,n) times (-n,n).
$$
Clearly, $G_n subset G_n+1$ for each $n$ and these form an open cover of $mathbbR^2$. In particular, they cover $K$. By compactness, we can cover $K$ by finitely many of these $G_n$. Since the $G_n$ are increasing, we can therefore find $N in mathbbN$ such that $K subseteq G_N$. In this case,
$$
K subseteq G_n subset [-N,N] times [-N,N].
$$
Additional Note. As pointed out in the comments, be warned that not every compact subset of $mathbbR$ is an interval. By the Heine-Borel theorem, a set $K subseteq mathbbR$ is compact if and only if it is closed and bounded. However, there is no reason for $K$ to be either finite or an interval. For instance, the Cantor set $mathfrakC$ is compact but is neither countable nor an interval.
answered Mar 27 at 3:12
rolandcyprolandcyp
2,149422
$endgroup$
1
$begingroup$
My edit was for a trivial typo.
$endgroup$
– DanielWainfleet
Mar 27 at 4:47
$begingroup$
@DanielWainfleet Thank you!
$endgroup$
– rolandcyp
Mar 27 at 5:01
add a comment |
$begingroup$
There is $c>0$ such that $sqrtx^2+y^2 le c$ for all $(x,y) in K.$
Now let $(x,y) in K.$.
Then $|x|=sqrtx^2 le sqrtx^2+y^2 le c$ and $|y|=sqrty^2 le sqrtx^2+y^2 le c$.
Thus $(x,y) in [-c,c] times [-c,c].$ This gives $K subseteq [-c,c] times [-c,c].$
Remark: all we need is that $K$ is bounded !
answered Mar 27 at 6:09
FredFred
48.6k11849
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the family $left G_n right_n geqslant 1$ where each $G_n$ is the open subset of $mathbbR^2$ given by
$$
G_n := (-n,n) times (-n,n).
$$
Clearly, $G_n subset G_n+1$ for each $n$ and these form an open cover of $mathbbR^2$. In particular, they cover $K$. By compactness, we can cover $K$ by finitely many of these $G_n$. Since the $G_n$ are increasing, we can therefore find $N in mathbbN$ such that $K subseteq G_N$. In this case,
$$
K subseteq G_n subset [-N,N] times [-N,N].
$$
Additional Note. As pointed out in the comments, be warned that not every compact subset of $mathbbR$ is an interval. By the Heine-Borel theorem, a set $K subseteq mathbbR$ is compact if and only if it is closed and bounded. However, there is no reason for $K$ to be either finite or an interval. For instance, the Cantor set $mathfrakC$ is compact but is neither countable nor an interval.
answered Mar 27 at 3:12
rolandcyprolandcyp
2,149422
$endgroup$
1
$begingroup$
My edit was for a trivial typo.
$endgroup$
– DanielWainfleet
Mar 27 at 4:47
$begingroup$
@DanielWainfleet Thank you!
$endgroup$
– rolandcyp
Mar 27 at 5:01
add a comment |
$begingroup$
Consider the family $left G_n right_n geqslant 1$ where each $G_n$ is the open subset of $mathbbR^2$ given by
$$
G_n := (-n,n) times (-n,n).
$$
Clearly, $G_n subset G_n+1$ for each $n$ and these form an open cover of $mathbbR^2$. In particular, they cover $K$. By compactness, we can cover $K$ by finitely many of these $G_n$. Since the $G_n$ are increasing, we can therefore find $N in mathbbN$ such that $K subseteq G_N$. In this case,
$$
K subseteq G_n subset [-N,N] times [-N,N].
$$
Additional Note. As pointed out in the comments, be warned that not every compact subset of $mathbbR$ is an interval. By the Heine-Borel theorem, a set $K subseteq mathbbR$ is compact if and only if it is closed and bounded. However, there is no reason for $K$ to be either finite or an interval. For instance, the Cantor set $mathfrakC$ is compact but is neither countable nor an interval.
answered Mar 27 at 3:12
rolandcyprolandcyp
2,149422
$endgroup$
1
$begingroup$
My edit was for a trivial typo.
$endgroup$
– DanielWainfleet
Mar 27 at 4:47
$begingroup$
@DanielWainfleet Thank you!
$endgroup$
– rolandcyp
Mar 27 at 5:01
add a comment |
$begingroup$
Consider the family $left G_n right_n geqslant 1$ where each $G_n$ is the open subset of $mathbbR^2$ given by
$$
G_n := (-n,n) times (-n,n).
$$
Clearly, $G_n subset G_n+1$ for each $n$ and these form an open cover of $mathbbR^2$. In particular, they cover $K$. By compactness, we can cover $K$ by finitely many of these $G_n$. Since the $G_n$ are increasing, we can therefore find $N in mathbbN$ such that $K subseteq G_N$. In this case,
$$
K subseteq G_n subset [-N,N] times [-N,N].
$$
Additional Note. As pointed out in the comments, be warned that not every compact subset of $mathbbR$ is an interval. By the Heine-Borel theorem, a set $K subseteq mathbbR$ is compact if and only if it is closed and bounded. However, there is no reason for $K$ to be either finite or an interval. For instance, the Cantor set $mathfrakC$ is compact but is neither countable nor an interval.
answered Mar 27 at 3:12
rolandcyprolandcyp
2,149422
$endgroup$
Consider the family $left G_n right_n geqslant 1$ where each $G_n$ is the open subset of $mathbbR^2$ given by
$$
G_n := (-n,n) times (-n,n).
$$
Clearly, $G_n subset G_n+1$ for each $n$ and these form an open cover of $mathbbR^2$. In particular, they cover $K$. By compactness, we can cover $K$ by finitely many of these $G_n$. Since the $G_n$ are increasing, we can therefore find $N in mathbbN$ such that $K subseteq G_N$. In this case,
$$
K subseteq G_n subset [-N,N] times [-N,N].
$$
Additional Note. As pointed out in the comments, be warned that not every compact subset of $mathbbR$ is an interval. By the Heine-Borel theorem, a set $K subseteq mathbbR$ is compact if and only if it is closed and bounded. However, there is no reason for $K$ to be either finite or an interval. For instance, the Cantor set $mathfrakC$ is compact but is neither countable nor an interval.
answered Mar 27 at 3:12
rolandcyprolandcyp
2,149422
edited Mar 27 at 5:03
answered Mar 27 at 3:12
rolandcyprolandcyp
2,149422
answered Mar 27 at 3:12
rolandcyprolandcyp
2,149422
answered Mar 27 at 3:12
rolandcyprolandcyp
2,149422
2,149422
1
$begingroup$
My edit was for a trivial typo.
$endgroup$
– DanielWainfleet
Mar 27 at 4:47
$begingroup$
@DanielWainfleet Thank you!
$endgroup$
– rolandcyp
Mar 27 at 5:01
add a comment |
1
$begingroup$
My edit was for a trivial typo.
$endgroup$
– DanielWainfleet
Mar 27 at 4:47
$begingroup$
@DanielWainfleet Thank you!
$endgroup$
– rolandcyp
Mar 27 at 5:01
1
1
$begingroup$
My edit was for a trivial typo.
$endgroup$
– DanielWainfleet
Mar 27 at 4:47
$begingroup$
My edit was for a trivial typo.
$endgroup$
– DanielWainfleet
Mar 27 at 4:47
$begingroup$
@DanielWainfleet Thank you!
$endgroup$
– rolandcyp
Mar 27 at 5:01
$begingroup$
@DanielWainfleet Thank you!
$endgroup$
– rolandcyp
Mar 27 at 5:01
add a comment |
$begingroup$
There is $c>0$ such that $sqrtx^2+y^2 le c$ for all $(x,y) in K.$
Now let $(x,y) in K.$.
Then $|x|=sqrtx^2 le sqrtx^2+y^2 le c$ and $|y|=sqrty^2 le sqrtx^2+y^2 le c$.
Thus $(x,y) in [-c,c] times [-c,c].$ This gives $K subseteq [-c,c] times [-c,c].$
Remark: all we need is that $K$ is bounded !
answered Mar 27 at 6:09
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
There is $c>0$ such that $sqrtx^2+y^2 le c$ for all $(x,y) in K.$
Now let $(x,y) in K.$.
Then $|x|=sqrtx^2 le sqrtx^2+y^2 le c$ and $|y|=sqrty^2 le sqrtx^2+y^2 le c$.
Thus $(x,y) in [-c,c] times [-c,c].$ This gives $K subseteq [-c,c] times [-c,c].$
Remark: all we need is that $K$ is bounded !
answered Mar 27 at 6:09
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
There is $c>0$ such that $sqrtx^2+y^2 le c$ for all $(x,y) in K.$
Now let $(x,y) in K.$.
Then $|x|=sqrtx^2 le sqrtx^2+y^2 le c$ and $|y|=sqrty^2 le sqrtx^2+y^2 le c$.
Thus $(x,y) in [-c,c] times [-c,c].$ This gives $K subseteq [-c,c] times [-c,c].$
Remark: all we need is that $K$ is bounded !
answered Mar 27 at 6:09
FredFred
48.6k11849
$endgroup$
There is $c>0$ such that $sqrtx^2+y^2 le c$ for all $(x,y) in K.$
Now let $(x,y) in K.$.
Then $|x|=sqrtx^2 le sqrtx^2+y^2 le c$ and $|y|=sqrty^2 le sqrtx^2+y^2 le c$.
Thus $(x,y) in [-c,c] times [-c,c].$ This gives $K subseteq [-c,c] times [-c,c].$
Remark: all we need is that $K$ is bounded !
answered Mar 27 at 6:09
FredFred
48.6k11849
answered Mar 27 at 6:09
FredFred
48.6k11849
answered Mar 27 at 6:09
FredFred
48.6k11849
answered Mar 27 at 6:09
FredFred
48.6k11849
48.6k11849
add a comment |
add a comment |
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1
$begingroup$
Hint: Show $K$ must be bounded.
$endgroup$
– Theo Bendit
Mar 27 at 3:11
1
$begingroup$
There are exotic compact sets in the real line, not only intervals...
$endgroup$
– Eduardo Longa
Mar 27 at 3:13
4
$begingroup$
You have the wrong typo in the title. The correct typo for subset is sunset.
$endgroup$
– DanielWainfleet
Mar 27 at 4:48