Suppose $f:[0,1]times[0,1]mapsto X$ is a continuous function. Show that $[0,1]times[0,1]$ can partitioned into rectangles s.t $f(R_i)subseteq U_k$Proving that sequentially compact spaces are compact.Relatively compact subsets of a manifold.Proof or definition of compactness in lecture notes?Prove that a metrizable space is countably compact iff it is compact.Is $mathbbR$ compact under the co-countable and co-finite topologies?Interval $[0,1]$ is neither compact nor connected in the Sorgenfrey line.Countably Compact Equivalent to Nested Sequence PropertyShow that a continuous function on a compact set satisfies this bound.Open Covers of Unit Balls in Different SpacesLimit point compact uniform space
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Suppose $f:[0,1]times[0,1]mapsto X$ is a continuous function. Show that $[0,1]times[0,1]$ can partitioned into rectangles s.t $f(R_i)subseteq U_k$
Proving that sequentially compact spaces are compact.Relatively compact subsets of a manifold.Proof or definition of compactness in lecture notes?Prove that a metrizable space is countably compact iff it is compact.Is $mathbbR$ compact under the co-countable and co-finite topologies?Interval $[0,1]$ is neither compact nor connected in the Sorgenfrey line.Countably Compact Equivalent to Nested Sequence PropertyShow that a continuous function on a compact set satisfies this bound.Open Covers of Unit Balls in Different SpacesLimit point compact uniform space
$begingroup$
Suppose $f:[0,1]times[0,1]mapsto X$, is a continuous function where $X$ compact and connected subset of $mathbbR^n$. Show that $[0,1]times[0,1]$ can partitioned into rectangles $R_i$ such that $f(R_i)subseteq U_i$ where $U_iin C$ a cover for $X$.
My attempt at a proof:
Choose a cover for $X$, then it has a finite subcover.
Suppose not, then for any rectangles with side lengths $xin mathbbN$, $f(R_i)notsubseteq U_i$ for any $U_iin C$. I believe I want to take rectangles of side lengths, $frac1n$ and then make a sequence of balls of radius $frac1m$, for $min mathbbN$ and then use sequential compactness to get the image can be contained in a single $U_i$.
general-topology compactness cauchy-sequences
asked Mar 21 at 23:28
AColoredReptileAColoredReptile
401210
$endgroup$
add a comment |
$begingroup$
Suppose $f:[0,1]times[0,1]mapsto X$, is a continuous function where $X$ compact and connected subset of $mathbbR^n$. Show that $[0,1]times[0,1]$ can partitioned into rectangles $R_i$ such that $f(R_i)subseteq U_i$ where $U_iin C$ a cover for $X$.
My attempt at a proof:
Choose a cover for $X$, then it has a finite subcover.
Suppose not, then for any rectangles with side lengths $xin mathbbN$, $f(R_i)notsubseteq U_i$ for any $U_iin C$. I believe I want to take rectangles of side lengths, $frac1n$ and then make a sequence of balls of radius $frac1m$, for $min mathbbN$ and then use sequential compactness to get the image can be contained in a single $U_i$.
general-topology compactness cauchy-sequences
asked Mar 21 at 23:28
AColoredReptileAColoredReptile
401210
$endgroup$
$begingroup$
Oh right so this definitely doesnt work.
$endgroup$
– AColoredReptile
Mar 21 at 23:45
$begingroup$
Partition R = [0,1]×[0.1] into R . Thus f(R) subset open X and X is an open cover of X.
$endgroup$
– William Elliot
Mar 22 at 1:26
$begingroup$
Do you know the Lebesgue Number Lemma?
$endgroup$
– Lee Mosher
Mar 22 at 2:02
$begingroup$
@Lee Mosher Yes, the proof I know of it is similar to the one I am trying to do, but I'm not sure how to make it work for the image of a function.
$endgroup$
– AColoredReptile
Mar 22 at 2:34
add a comment |
$begingroup$
Suppose $f:[0,1]times[0,1]mapsto X$, is a continuous function where $X$ compact and connected subset of $mathbbR^n$. Show that $[0,1]times[0,1]$ can partitioned into rectangles $R_i$ such that $f(R_i)subseteq U_i$ where $U_iin C$ a cover for $X$.
My attempt at a proof:
Choose a cover for $X$, then it has a finite subcover.
Suppose not, then for any rectangles with side lengths $xin mathbbN$, $f(R_i)notsubseteq U_i$ for any $U_iin C$. I believe I want to take rectangles of side lengths, $frac1n$ and then make a sequence of balls of radius $frac1m$, for $min mathbbN$ and then use sequential compactness to get the image can be contained in a single $U_i$.
general-topology compactness cauchy-sequences
asked Mar 21 at 23:28
AColoredReptileAColoredReptile
401210
$endgroup$
Suppose $f:[0,1]times[0,1]mapsto X$, is a continuous function where $X$ compact and connected subset of $mathbbR^n$. Show that $[0,1]times[0,1]$ can partitioned into rectangles $R_i$ such that $f(R_i)subseteq U_i$ where $U_iin C$ a cover for $X$.
My attempt at a proof:
Choose a cover for $X$, then it has a finite subcover.
Suppose not, then for any rectangles with side lengths $xin mathbbN$, $f(R_i)notsubseteq U_i$ for any $U_iin C$. I believe I want to take rectangles of side lengths, $frac1n$ and then make a sequence of balls of radius $frac1m$, for $min mathbbN$ and then use sequential compactness to get the image can be contained in a single $U_i$.
general-topology compactness cauchy-sequences
general-topology compactness cauchy-sequences
asked Mar 21 at 23:28
AColoredReptileAColoredReptile
401210
asked Mar 21 at 23:28
AColoredReptileAColoredReptile
401210
edited Mar 21 at 23:57
AColoredReptile
asked Mar 21 at 23:28
AColoredReptileAColoredReptile
401210
asked Mar 21 at 23:28
AColoredReptileAColoredReptile
401210
asked Mar 21 at 23:28
AColoredReptileAColoredReptile
401210
401210
$begingroup$
Oh right so this definitely doesnt work.
$endgroup$
– AColoredReptile
Mar 21 at 23:45
$begingroup$
Partition R = [0,1]×[0.1] into R . Thus f(R) subset open X and X is an open cover of X.
$endgroup$
– William Elliot
Mar 22 at 1:26
$begingroup$
Do you know the Lebesgue Number Lemma?
$endgroup$
– Lee Mosher
Mar 22 at 2:02
$begingroup$
@Lee Mosher Yes, the proof I know of it is similar to the one I am trying to do, but I'm not sure how to make it work for the image of a function.
$endgroup$
– AColoredReptile
Mar 22 at 2:34
add a comment |
$begingroup$
Oh right so this definitely doesnt work.
$endgroup$
– AColoredReptile
Mar 21 at 23:45
$begingroup$
Partition R = [0,1]×[0.1] into R . Thus f(R) subset open X and X is an open cover of X.
$endgroup$
– William Elliot
Mar 22 at 1:26
$begingroup$
Do you know the Lebesgue Number Lemma?
$endgroup$
– Lee Mosher
Mar 22 at 2:02
$begingroup$
@Lee Mosher Yes, the proof I know of it is similar to the one I am trying to do, but I'm not sure how to make it work for the image of a function.
$endgroup$
– AColoredReptile
Mar 22 at 2:34
$begingroup$
Oh right so this definitely doesnt work.
$endgroup$
– AColoredReptile
Mar 21 at 23:45
$begingroup$
Oh right so this definitely doesnt work.
$endgroup$
– AColoredReptile
Mar 21 at 23:45
$begingroup$
Partition R = [0,1]×[0.1] into R . Thus f(R) subset open X and X is an open cover of X.
$endgroup$
– William Elliot
Mar 22 at 1:26
$begingroup$
Partition R = [0,1]×[0.1] into R . Thus f(R) subset open X and X is an open cover of X.
$endgroup$
– William Elliot
Mar 22 at 1:26
$begingroup$
Do you know the Lebesgue Number Lemma?
$endgroup$
– Lee Mosher
Mar 22 at 2:02
$begingroup$
Do you know the Lebesgue Number Lemma?
$endgroup$
– Lee Mosher
Mar 22 at 2:02
$begingroup$
@Lee Mosher Yes, the proof I know of it is similar to the one I am trying to do, but I'm not sure how to make it work for the image of a function.
$endgroup$
– AColoredReptile
Mar 22 at 2:34
$begingroup$
@Lee Mosher Yes, the proof I know of it is similar to the one I am trying to do, but I'm not sure how to make it work for the image of a function.
$endgroup$
– AColoredReptile
Mar 22 at 2:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The easiest thing is to simply apply the Lebesgue Number Lemma to the open covering of $X$ that you get by pulling back the given open covering of $Y$, namely
$$cal V = f^-1(U_i) mid U_i in C
$$
If you apply the lemma you get a Lebesgue number $lambda>0$. Then subdivide $[0,1] times [0,1]$ into rectangles whose diagonal length is $<lambda$. Each little rectangle $R$ will be contained in some $f^-1(U_i) in calV$, and so
$f(R) subset U_i$.
answered Mar 22 at 2:58
Lee MosherLee Mosher
51.7k33889
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
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votes
$begingroup$
The easiest thing is to simply apply the Lebesgue Number Lemma to the open covering of $X$ that you get by pulling back the given open covering of $Y$, namely
$$cal V = f^-1(U_i) mid U_i in C
$$
If you apply the lemma you get a Lebesgue number $lambda>0$. Then subdivide $[0,1] times [0,1]$ into rectangles whose diagonal length is $<lambda$. Each little rectangle $R$ will be contained in some $f^-1(U_i) in calV$, and so
$f(R) subset U_i$.
answered Mar 22 at 2:58
Lee MosherLee Mosher
51.7k33889
$endgroup$
add a comment |
$begingroup$
The easiest thing is to simply apply the Lebesgue Number Lemma to the open covering of $X$ that you get by pulling back the given open covering of $Y$, namely
$$cal V = f^-1(U_i) mid U_i in C
$$
If you apply the lemma you get a Lebesgue number $lambda>0$. Then subdivide $[0,1] times [0,1]$ into rectangles whose diagonal length is $<lambda$. Each little rectangle $R$ will be contained in some $f^-1(U_i) in calV$, and so
$f(R) subset U_i$.
answered Mar 22 at 2:58
Lee MosherLee Mosher
51.7k33889
$endgroup$
add a comment |
$begingroup$
The easiest thing is to simply apply the Lebesgue Number Lemma to the open covering of $X$ that you get by pulling back the given open covering of $Y$, namely
$$cal V = f^-1(U_i) mid U_i in C
$$
If you apply the lemma you get a Lebesgue number $lambda>0$. Then subdivide $[0,1] times [0,1]$ into rectangles whose diagonal length is $<lambda$. Each little rectangle $R$ will be contained in some $f^-1(U_i) in calV$, and so
$f(R) subset U_i$.
answered Mar 22 at 2:58
Lee MosherLee Mosher
51.7k33889
$endgroup$
The easiest thing is to simply apply the Lebesgue Number Lemma to the open covering of $X$ that you get by pulling back the given open covering of $Y$, namely
$$cal V = f^-1(U_i) mid U_i in C
$$
If you apply the lemma you get a Lebesgue number $lambda>0$. Then subdivide $[0,1] times [0,1]$ into rectangles whose diagonal length is $<lambda$. Each little rectangle $R$ will be contained in some $f^-1(U_i) in calV$, and so
$f(R) subset U_i$.
answered Mar 22 at 2:58
Lee MosherLee Mosher
51.7k33889
edited Mar 22 at 14:07
answered Mar 22 at 2:58
Lee MosherLee Mosher
51.7k33889
answered Mar 22 at 2:58
Lee MosherLee Mosher
51.7k33889
answered Mar 22 at 2:58
Lee MosherLee Mosher
51.7k33889
51.7k33889
add a comment |
add a comment |
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$begingroup$
Oh right so this definitely doesnt work.
$endgroup$
– AColoredReptile
Mar 21 at 23:45
$begingroup$
Partition R = [0,1]×[0.1] into R . Thus f(R) subset open X and X is an open cover of X.
$endgroup$
– William Elliot
Mar 22 at 1:26
$begingroup$
Do you know the Lebesgue Number Lemma?
$endgroup$
– Lee Mosher
Mar 22 at 2:02
$begingroup$
@Lee Mosher Yes, the proof I know of it is similar to the one I am trying to do, but I'm not sure how to make it work for the image of a function.
$endgroup$
– AColoredReptile
Mar 22 at 2:34