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Showing $f:[0,1] rightarrow mathbbR$ has finitely many isolated zeros
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We say that $x_0 in mathbbR$ is an isolated zero of $mathbbR$-valued function $f$ $f(x_0) = 0$ and there exists an open set containing $x_0$ that contains no other zeros of $f$. Show that if $f : [0,1] rightarrow mathbbR$ is continuous and all of its zeros are isolated, then it only has finitely many isolated zeros.
Now I think I am on the right track here, but I need some help finishing this off. Let $epsilon > 0$ and set $K : = [- epsilon, epsilon]$. Then $f^-1(K)$ contains all of the zeros of $f$ by construction and is closed since $f$ is continuous and $K$ is closed. Moreover, since $[0,1]$ is compact and $f^-1(K)$ is closed, we know that $f^-1(K)$ is compact.
From here, I think that I should say that if there were infinitely many isolated zeros then we could cook up an open cover without a finite subcover thereby contradicting compactness. To this end, let $ x_alpha _alpha in A$ be the zeros of $f$ (which are all isolated by assumption) and let $U_alpha _alpha in A$ be a corresponding collection of open sets such that $x_alpha in U_alpha$ and $x_beta notin U_alpha$ for $alpha neq beta$, where $A$ is some infinite index set. If we let $U subseteq [0,1]$ be an open set such that $f^-1(K) setminus bigcup_alpha in A U_alpha subseteq U$ and $U cap bigcup_alpha in A U_alpha = emptyset$. Then $U cup bigcup_alpha in A U_alpha$ constitutes an open cover $f^-1(K)$ and thus we may refine it to a finite subcover. But any finite subcover will not contain infinitely many $U_alpha$ and $x_alpha notin U_beta$ for $alpha neq beta$ and $x_alpha notin U$ by construction, which yields a contradiction.
Is this correct? Also is there a more direct way of showing this because I definitely feel like my argument could be improved upon.
real-analysis general-topology continuity compactness
asked Jul 28 at 21:59
Nala Ball
555
add a comment |Â
up vote
3
down vote
favorite
We say that $x_0 in mathbbR$ is an isolated zero of $mathbbR$-valued function $f$ $f(x_0) = 0$ and there exists an open set containing $x_0$ that contains no other zeros of $f$. Show that if $f : [0,1] rightarrow mathbbR$ is continuous and all of its zeros are isolated, then it only has finitely many isolated zeros.
Now I think I am on the right track here, but I need some help finishing this off. Let $epsilon > 0$ and set $K : = [- epsilon, epsilon]$. Then $f^-1(K)$ contains all of the zeros of $f$ by construction and is closed since $f$ is continuous and $K$ is closed. Moreover, since $[0,1]$ is compact and $f^-1(K)$ is closed, we know that $f^-1(K)$ is compact.
From here, I think that I should say that if there were infinitely many isolated zeros then we could cook up an open cover without a finite subcover thereby contradicting compactness. To this end, let $ x_alpha _alpha in A$ be the zeros of $f$ (which are all isolated by assumption) and let $U_alpha _alpha in A$ be a corresponding collection of open sets such that $x_alpha in U_alpha$ and $x_beta notin U_alpha$ for $alpha neq beta$, where $A$ is some infinite index set. If we let $U subseteq [0,1]$ be an open set such that $f^-1(K) setminus bigcup_alpha in A U_alpha subseteq U$ and $U cap bigcup_alpha in A U_alpha = emptyset$. Then $U cup bigcup_alpha in A U_alpha$ constitutes an open cover $f^-1(K)$ and thus we may refine it to a finite subcover. But any finite subcover will not contain infinitely many $U_alpha$ and $x_alpha notin U_beta$ for $alpha neq beta$ and $x_alpha notin U$ by construction, which yields a contradiction.
Is this correct? Also is there a more direct way of showing this because I definitely feel like my argument could be improved upon.
real-analysis general-topology continuity compactness
asked Jul 28 at 21:59
Nala Ball
555
1
Limit point compact and compact are the same, here.
– Benjamin Tighe
Jul 28 at 22:02
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
We say that $x_0 in mathbbR$ is an isolated zero of $mathbbR$-valued function $f$ $f(x_0) = 0$ and there exists an open set containing $x_0$ that contains no other zeros of $f$. Show that if $f : [0,1] rightarrow mathbbR$ is continuous and all of its zeros are isolated, then it only has finitely many isolated zeros.
Now I think I am on the right track here, but I need some help finishing this off. Let $epsilon > 0$ and set $K : = [- epsilon, epsilon]$. Then $f^-1(K)$ contains all of the zeros of $f$ by construction and is closed since $f$ is continuous and $K$ is closed. Moreover, since $[0,1]$ is compact and $f^-1(K)$ is closed, we know that $f^-1(K)$ is compact.
From here, I think that I should say that if there were infinitely many isolated zeros then we could cook up an open cover without a finite subcover thereby contradicting compactness. To this end, let $ x_alpha _alpha in A$ be the zeros of $f$ (which are all isolated by assumption) and let $U_alpha _alpha in A$ be a corresponding collection of open sets such that $x_alpha in U_alpha$ and $x_beta notin U_alpha$ for $alpha neq beta$, where $A$ is some infinite index set. If we let $U subseteq [0,1]$ be an open set such that $f^-1(K) setminus bigcup_alpha in A U_alpha subseteq U$ and $U cap bigcup_alpha in A U_alpha = emptyset$. Then $U cup bigcup_alpha in A U_alpha$ constitutes an open cover $f^-1(K)$ and thus we may refine it to a finite subcover. But any finite subcover will not contain infinitely many $U_alpha$ and $x_alpha notin U_beta$ for $alpha neq beta$ and $x_alpha notin U$ by construction, which yields a contradiction.
Is this correct? Also is there a more direct way of showing this because I definitely feel like my argument could be improved upon.
real-analysis general-topology continuity compactness
asked Jul 28 at 21:59
Nala Ball
555
We say that $x_0 in mathbbR$ is an isolated zero of $mathbbR$-valued function $f$ $f(x_0) = 0$ and there exists an open set containing $x_0$ that contains no other zeros of $f$. Show that if $f : [0,1] rightarrow mathbbR$ is continuous and all of its zeros are isolated, then it only has finitely many isolated zeros.
Now I think I am on the right track here, but I need some help finishing this off. Let $epsilon > 0$ and set $K : = [- epsilon, epsilon]$. Then $f^-1(K)$ contains all of the zeros of $f$ by construction and is closed since $f$ is continuous and $K$ is closed. Moreover, since $[0,1]$ is compact and $f^-1(K)$ is closed, we know that $f^-1(K)$ is compact.
From here, I think that I should say that if there were infinitely many isolated zeros then we could cook up an open cover without a finite subcover thereby contradicting compactness. To this end, let $ x_alpha _alpha in A$ be the zeros of $f$ (which are all isolated by assumption) and let $U_alpha _alpha in A$ be a corresponding collection of open sets such that $x_alpha in U_alpha$ and $x_beta notin U_alpha$ for $alpha neq beta$, where $A$ is some infinite index set. If we let $U subseteq [0,1]$ be an open set such that $f^-1(K) setminus bigcup_alpha in A U_alpha subseteq U$ and $U cap bigcup_alpha in A U_alpha = emptyset$. Then $U cup bigcup_alpha in A U_alpha$ constitutes an open cover $f^-1(K)$ and thus we may refine it to a finite subcover. But any finite subcover will not contain infinitely many $U_alpha$ and $x_alpha notin U_beta$ for $alpha neq beta$ and $x_alpha notin U$ by construction, which yields a contradiction.
Is this correct? Also is there a more direct way of showing this because I definitely feel like my argument could be improved upon.
real-analysis general-topology continuity compactness
asked Jul 28 at 21:59
Nala Ball
555
asked Jul 28 at 21:59
Nala Ball
555
asked Jul 28 at 21:59
Nala Ball
555
asked Jul 28 at 21:59
Nala Ball
555
555
1
Limit point compact and compact are the same, here.
– Benjamin Tighe
Jul 28 at 22:02
add a comment |Â
1
Limit point compact and compact are the same, here.
– Benjamin Tighe
Jul 28 at 22:02
1
1
Limit point compact and compact are the same, here.
– Benjamin Tighe
Jul 28 at 22:02
Limit point compact and compact are the same, here.
– Benjamin Tighe
Jul 28 at 22:02
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
I didn't find any error, but there is a simpler approach. Let $A$ be the set of non-zeros of $f$. For each zero $z$ of $f$, let $U_z$ be a neighborhood of $z$ with no other zero. So, if $Z$ is the set of zeros of $f$, $A$ together with the $U_z$'s form an open cover of $[0,1]$ and each element of this open cover has a zero, at most. So, if $Z$ was infinite, there would be no finite subcover.
answered Jul 28 at 22:08
José Carlos Santos
112k1696173
Terrific and way simpler. I was looking for something like this. Thank you.
– Nala Ball
Jul 28 at 22:13
I'm glad I could help.
– José Carlos Santos
Jul 28 at 22:18
add a comment |Â
up vote
2
down vote
If the set of zeroes say $A$ is infinite then being bounded $A$ has a limit point in $[0,1]$ say $c$. Now we must have $f(c) =0$ and clearly $c$ would not be an isolated zero. To see why $f(c) =0$ just assume $f(c) neq 0$ and by continuity there is a neighborhood of $c$ where $f$ is non-zero. Then the is neighborhood does not contain any point of $A$ contradicting that $c$ is a limit point of $A$.
answered Jul 29 at 3:24
Paramanand Singh
45k553142
add a comment |Â
up vote
1
down vote
If the set of zeros was not finite, then there would be some sequence of distinct zeros
$x_n$ such that $x_n to x$. By continuity, $f(x) = 0$, which contradicts the
assumption that the zeros are isolated.
answered Jul 29 at 2:22
copper.hat
122k557156
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
I didn't find any error, but there is a simpler approach. Let $A$ be the set of non-zeros of $f$. For each zero $z$ of $f$, let $U_z$ be a neighborhood of $z$ with no other zero. So, if $Z$ is the set of zeros of $f$, $A$ together with the $U_z$'s form an open cover of $[0,1]$ and each element of this open cover has a zero, at most. So, if $Z$ was infinite, there would be no finite subcover.
answered Jul 28 at 22:08
José Carlos Santos
112k1696173
Terrific and way simpler. I was looking for something like this. Thank you.
– Nala Ball
Jul 28 at 22:13
I'm glad I could help.
– José Carlos Santos
Jul 28 at 22:18
add a comment |Â
up vote
5
down vote
accepted
I didn't find any error, but there is a simpler approach. Let $A$ be the set of non-zeros of $f$. For each zero $z$ of $f$, let $U_z$ be a neighborhood of $z$ with no other zero. So, if $Z$ is the set of zeros of $f$, $A$ together with the $U_z$'s form an open cover of $[0,1]$ and each element of this open cover has a zero, at most. So, if $Z$ was infinite, there would be no finite subcover.
answered Jul 28 at 22:08
José Carlos Santos
112k1696173
Terrific and way simpler. I was looking for something like this. Thank you.
– Nala Ball
Jul 28 at 22:13
I'm glad I could help.
– José Carlos Santos
Jul 28 at 22:18
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
I didn't find any error, but there is a simpler approach. Let $A$ be the set of non-zeros of $f$. For each zero $z$ of $f$, let $U_z$ be a neighborhood of $z$ with no other zero. So, if $Z$ is the set of zeros of $f$, $A$ together with the $U_z$'s form an open cover of $[0,1]$ and each element of this open cover has a zero, at most. So, if $Z$ was infinite, there would be no finite subcover.
answered Jul 28 at 22:08
José Carlos Santos
112k1696173
I didn't find any error, but there is a simpler approach. Let $A$ be the set of non-zeros of $f$. For each zero $z$ of $f$, let $U_z$ be a neighborhood of $z$ with no other zero. So, if $Z$ is the set of zeros of $f$, $A$ together with the $U_z$'s form an open cover of $[0,1]$ and each element of this open cover has a zero, at most. So, if $Z$ was infinite, there would be no finite subcover.
answered Jul 28 at 22:08
José Carlos Santos
112k1696173
answered Jul 28 at 22:08
José Carlos Santos
112k1696173
answered Jul 28 at 22:08
José Carlos Santos
112k1696173
answered Jul 28 at 22:08
José Carlos Santos
112k1696173
112k1696173
Terrific and way simpler. I was looking for something like this. Thank you.
– Nala Ball
Jul 28 at 22:13
I'm glad I could help.
– José Carlos Santos
Jul 28 at 22:18
add a comment |Â
Terrific and way simpler. I was looking for something like this. Thank you.
– Nala Ball
Jul 28 at 22:13
I'm glad I could help.
– José Carlos Santos
Jul 28 at 22:18
Terrific and way simpler. I was looking for something like this. Thank you.
– Nala Ball
Jul 28 at 22:13
Terrific and way simpler. I was looking for something like this. Thank you.
– Nala Ball
Jul 28 at 22:13
I'm glad I could help.
– José Carlos Santos
Jul 28 at 22:18
I'm glad I could help.
– José Carlos Santos
Jul 28 at 22:18
add a comment |Â
up vote
2
down vote
If the set of zeroes say $A$ is infinite then being bounded $A$ has a limit point in $[0,1]$ say $c$. Now we must have $f(c) =0$ and clearly $c$ would not be an isolated zero. To see why $f(c) =0$ just assume $f(c) neq 0$ and by continuity there is a neighborhood of $c$ where $f$ is non-zero. Then the is neighborhood does not contain any point of $A$ contradicting that $c$ is a limit point of $A$.
answered Jul 29 at 3:24
Paramanand Singh
45k553142
add a comment |Â
up vote
2
down vote
If the set of zeroes say $A$ is infinite then being bounded $A$ has a limit point in $[0,1]$ say $c$. Now we must have $f(c) =0$ and clearly $c$ would not be an isolated zero. To see why $f(c) =0$ just assume $f(c) neq 0$ and by continuity there is a neighborhood of $c$ where $f$ is non-zero. Then the is neighborhood does not contain any point of $A$ contradicting that $c$ is a limit point of $A$.
answered Jul 29 at 3:24
Paramanand Singh
45k553142
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If the set of zeroes say $A$ is infinite then being bounded $A$ has a limit point in $[0,1]$ say $c$. Now we must have $f(c) =0$ and clearly $c$ would not be an isolated zero. To see why $f(c) =0$ just assume $f(c) neq 0$ and by continuity there is a neighborhood of $c$ where $f$ is non-zero. Then the is neighborhood does not contain any point of $A$ contradicting that $c$ is a limit point of $A$.
answered Jul 29 at 3:24
Paramanand Singh
45k553142
If the set of zeroes say $A$ is infinite then being bounded $A$ has a limit point in $[0,1]$ say $c$. Now we must have $f(c) =0$ and clearly $c$ would not be an isolated zero. To see why $f(c) =0$ just assume $f(c) neq 0$ and by continuity there is a neighborhood of $c$ where $f$ is non-zero. Then the is neighborhood does not contain any point of $A$ contradicting that $c$ is a limit point of $A$.
answered Jul 29 at 3:24
Paramanand Singh
45k553142
answered Jul 29 at 3:24
Paramanand Singh
45k553142
answered Jul 29 at 3:24
Paramanand Singh
45k553142
answered Jul 29 at 3:24
Paramanand Singh
45k553142
45k553142
add a comment |Â
add a comment |Â
up vote
1
down vote
If the set of zeros was not finite, then there would be some sequence of distinct zeros
$x_n$ such that $x_n to x$. By continuity, $f(x) = 0$, which contradicts the
assumption that the zeros are isolated.
answered Jul 29 at 2:22
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
If the set of zeros was not finite, then there would be some sequence of distinct zeros
$x_n$ such that $x_n to x$. By continuity, $f(x) = 0$, which contradicts the
assumption that the zeros are isolated.
answered Jul 29 at 2:22
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If the set of zeros was not finite, then there would be some sequence of distinct zeros
$x_n$ such that $x_n to x$. By continuity, $f(x) = 0$, which contradicts the
assumption that the zeros are isolated.
answered Jul 29 at 2:22
copper.hat
122k557156
If the set of zeros was not finite, then there would be some sequence of distinct zeros
$x_n$ such that $x_n to x$. By continuity, $f(x) = 0$, which contradicts the
assumption that the zeros are isolated.
answered Jul 29 at 2:22
copper.hat
122k557156
answered Jul 29 at 2:22
copper.hat
122k557156
answered Jul 29 at 2:22
copper.hat
122k557156
answered Jul 29 at 2:22
copper.hat
122k557156
122k557156
add a comment |Â
add a comment |Â
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1
Limit point compact and compact are the same, here.
– Benjamin Tighe
Jul 28 at 22:02