Mixing
Prove that there is a first time the function is zero
$begingroup$
Let $f : [0, 1] to mathbb R$ be continuous and satisfy $f(0) = −1$, $f(1) = 1$.
Prove that there is a first time at which $ f $ is zero: that is, a number $s in (0, 1)$ with
$f(s) = 0$ but $f(t) not= 0$ if $t < s$.
I'm thinking of using the least upper bound property on the set and somehow proving it is $ 0 $ at the value?
Any hints?
real-analysis continuity
edited Jan 28 at 3:36
user549397
1,5761418
asked Jan 28 at 3:14
Dis-integratingDis-integrating
1,037526
$endgroup$
|
show 2 more comments
$begingroup$
Let $f : [0, 1] to mathbb R$ be continuous and satisfy $f(0) = −1$, $f(1) = 1$.
Prove that there is a first time at which $ f $ is zero: that is, a number $s in (0, 1)$ with
$f(s) = 0$ but $f(t) not= 0$ if $t < s$.
I'm thinking of using the least upper bound property on the set and somehow proving it is $ 0 $ at the value?
Any hints?
real-analysis continuity
edited Jan 28 at 3:36
user549397
1,5761418
asked Jan 28 at 3:14
Dis-integratingDis-integrating
1,037526
$endgroup$
$begingroup$
You should use the supremum to come up with a candidate for "the first time at which $f$ is zero". Can you think of such a candidate?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 3:20
$begingroup$
Intermediate Value Theorem tells us that there is at least one root. Now, can we prove that given this case, the function has at most countably many roots?
$endgroup$
– Aniruddha Deshmukh
Jan 28 at 3:20
$begingroup$
@AniruddhaDeshmukh no-one said anything about countably many roots and there's no reason to assume countablity. Consider $f(x) = 0$ for all $frac 14 le x le frac 34$ but $f(x) < 0$ for $x < frac 14$ and $f(x) > 0$ for $x >frac 34$. That has uncountably many roots but $x = frac 14$ is the least ("first") $x$ so that $f(x) = 0$.
$endgroup$
– fleablood
Jan 28 at 3:35
1
$begingroup$
Somebody asked exactly the same question few hours ago: math.stackexchange.com/questions/3090223/…
$endgroup$
– N. S.
Jan 28 at 3:48
1
$begingroup$
Possible duplicate of A function is continuous. Let $f : [0, 1] to mathbb R$ satisfy $f(0) = −1$, $f(1) = 1$. Prove that there is a first time at which f is zero
$endgroup$
– N. S.
Jan 28 at 3:48
|
show 2 more comments
$begingroup$
Let $f : [0, 1] to mathbb R$ be continuous and satisfy $f(0) = −1$, $f(1) = 1$.
Prove that there is a first time at which $ f $ is zero: that is, a number $s in (0, 1)$ with
$f(s) = 0$ but $f(t) not= 0$ if $t < s$.
I'm thinking of using the least upper bound property on the set and somehow proving it is $ 0 $ at the value?
Any hints?
real-analysis continuity
edited Jan 28 at 3:36
user549397
1,5761418
asked Jan 28 at 3:14
Dis-integratingDis-integrating
1,037526
$endgroup$
Let $f : [0, 1] to mathbb R$ be continuous and satisfy $f(0) = −1$, $f(1) = 1$.
Prove that there is a first time at which $ f $ is zero: that is, a number $s in (0, 1)$ with
$f(s) = 0$ but $f(t) not= 0$ if $t < s$.
I'm thinking of using the least upper bound property on the set and somehow proving it is $ 0 $ at the value?
Any hints?
real-analysis continuity
real-analysis continuity
edited Jan 28 at 3:36
user549397
1,5761418
asked Jan 28 at 3:14
Dis-integratingDis-integrating
1,037526
edited Jan 28 at 3:36
user549397
1,5761418
asked Jan 28 at 3:14
Dis-integratingDis-integrating
1,037526
edited Jan 28 at 3:36
user549397
1,5761418
edited Jan 28 at 3:36
user549397
1,5761418
edited Jan 28 at 3:36
user549397
1,5761418
1,5761418
asked Jan 28 at 3:14
Dis-integratingDis-integrating
1,037526
asked Jan 28 at 3:14
Dis-integratingDis-integrating
1,037526
asked Jan 28 at 3:14
Dis-integratingDis-integrating
1,037526
1,037526
$begingroup$
You should use the supremum to come up with a candidate for "the first time at which $f$ is zero". Can you think of such a candidate?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 3:20
$begingroup$
Intermediate Value Theorem tells us that there is at least one root. Now, can we prove that given this case, the function has at most countably many roots?
$endgroup$
– Aniruddha Deshmukh
Jan 28 at 3:20
$begingroup$
@AniruddhaDeshmukh no-one said anything about countably many roots and there's no reason to assume countablity. Consider $f(x) = 0$ for all $frac 14 le x le frac 34$ but $f(x) < 0$ for $x < frac 14$ and $f(x) > 0$ for $x >frac 34$. That has uncountably many roots but $x = frac 14$ is the least ("first") $x$ so that $f(x) = 0$.
$endgroup$
– fleablood
Jan 28 at 3:35
1
$begingroup$
Somebody asked exactly the same question few hours ago: math.stackexchange.com/questions/3090223/…
$endgroup$
– N. S.
Jan 28 at 3:48
1
$begingroup$
Possible duplicate of A function is continuous. Let $f : [0, 1] to mathbb R$ satisfy $f(0) = −1$, $f(1) = 1$. Prove that there is a first time at which f is zero
$endgroup$
– N. S.
Jan 28 at 3:48
|
show 2 more comments
$begingroup$
You should use the supremum to come up with a candidate for "the first time at which $f$ is zero". Can you think of such a candidate?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 3:20
$begingroup$
Intermediate Value Theorem tells us that there is at least one root. Now, can we prove that given this case, the function has at most countably many roots?
$endgroup$
– Aniruddha Deshmukh
Jan 28 at 3:20
$begingroup$
@AniruddhaDeshmukh no-one said anything about countably many roots and there's no reason to assume countablity. Consider $f(x) = 0$ for all $frac 14 le x le frac 34$ but $f(x) < 0$ for $x < frac 14$ and $f(x) > 0$ for $x >frac 34$. That has uncountably many roots but $x = frac 14$ is the least ("first") $x$ so that $f(x) = 0$.
$endgroup$
– fleablood
Jan 28 at 3:35
1
$begingroup$
Somebody asked exactly the same question few hours ago: math.stackexchange.com/questions/3090223/…
$endgroup$
– N. S.
Jan 28 at 3:48
1
$begingroup$
Possible duplicate of A function is continuous. Let $f : [0, 1] to mathbb R$ satisfy $f(0) = −1$, $f(1) = 1$. Prove that there is a first time at which f is zero
$endgroup$
– N. S.
Jan 28 at 3:48
$begingroup$
You should use the supremum to come up with a candidate for "the first time at which $f$ is zero". Can you think of such a candidate?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 3:20
$begingroup$
You should use the supremum to come up with a candidate for "the first time at which $f$ is zero". Can you think of such a candidate?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 28 at 3:20
$begingroup$
Intermediate Value Theorem tells us that there is at least one root. Now, can we prove that given this case, the function has at most countably many roots?
$endgroup$
– Aniruddha Deshmukh
Jan 28 at 3:20
$begingroup$
Intermediate Value Theorem tells us that there is at least one root. Now, can we prove that given this case, the function has at most countably many roots?
$endgroup$
– Aniruddha Deshmukh
Jan 28 at 3:20
$begingroup$
@AniruddhaDeshmukh no-one said anything about countably many roots and there's no reason to assume countablity. Consider $f(x) = 0$ for all $frac 14 le x le frac 34$ but $f(x) < 0$ for $x < frac 14$ and $f(x) > 0$ for $x >frac 34$. That has uncountably many roots but $x = frac 14$ is the least ("first") $x$ so that $f(x) = 0$.
$endgroup$
– fleablood
Jan 28 at 3:35
$begingroup$
@AniruddhaDeshmukh no-one said anything about countably many roots and there's no reason to assume countablity. Consider $f(x) = 0$ for all $frac 14 le x le frac 34$ but $f(x) < 0$ for $x < frac 14$ and $f(x) > 0$ for $x >frac 34$. That has uncountably many roots but $x = frac 14$ is the least ("first") $x$ so that $f(x) = 0$.
$endgroup$
– fleablood
Jan 28 at 3:35
1
1
$begingroup$
Somebody asked exactly the same question few hours ago: math.stackexchange.com/questions/3090223/…
$endgroup$
– N. S.
Jan 28 at 3:48
$begingroup$
Somebody asked exactly the same question few hours ago: math.stackexchange.com/questions/3090223/…
$endgroup$
– N. S.
Jan 28 at 3:48
1
1
$begingroup$
Possible duplicate of A function is continuous. Let $f : [0, 1] to mathbb R$ satisfy $f(0) = −1$, $f(1) = 1$. Prove that there is a first time at which f is zero
$endgroup$
– N. S.
Jan 28 at 3:48
$begingroup$
Possible duplicate of A function is continuous. Let $f : [0, 1] to mathbb R$ satisfy $f(0) = −1$, $f(1) = 1$. Prove that there is a first time at which f is zero
$endgroup$
– N. S.
Jan 28 at 3:48
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Suppose there is no first time. Then, for any $sin (0,1)$ such that $f(s)=0$ (why do you know that even one such $s$ exists?) there exists $s'<s$ such that $f(s')=0$ as well. We can keep doing this and get a decreasing sequence of points ${s_i}_{iinmathbb{N}}$. Does this sequence converge? How do convergent sequences behave with continuous functions?
answered Jan 28 at 3:22
Anubhav NanavatyAnubhav Nanavaty
714
$endgroup$
add a comment |
$begingroup$
We need the following fact: Since $f$ is continuous, the set $Z(f)={xin[0,1]:f(x)=0}$ is a closed set.
It is non-empty by intermediate value theorem. Then $s=inf Z(f)=min Z(f)$ is the first time at which $f$ is zero. It is clear $sin (0,1)$.
answered Jan 28 at 3:20
Eclipse SunEclipse Sun
7,8401438
$endgroup$
add a comment |
$begingroup$
For an argument from first principles, try defining
$$
S = {t in (0, 1) : f(x) < 0 text{ for all } x in [0, t) }.
$$
By the continuity of $f$ at $0$, $S$ is non-empty. By the continuity of $f$ at $1$, $S$ has an upper bound in $(0, 1)$. So $S$ has a least upper bound, $s in (0, 1)$. Now prove (i) $s in S$, (ii) $f(s) leqslant 0$, (iii) $f(s) geqslant 0$.
(I hope this is OK. You asked for a "hint", so I've left plenty of details to fill in. It's possible I've fooled myself!)
answered Jan 28 at 3:51
Calum GilhooleyCalum Gilhooley
5,119630
$endgroup$
add a comment |
$begingroup$
We know $f(c) = 0$ for some $c in [0,1]$ by the Intermediate Value Theorem.
So $S = {x|x in [0,1]; f(x) = 0}$ is not an empty set and it is bounded by $0, 1$. So $a=inf S$ exists.
What is $f(a)$.
Well $f(a) > 0$ is impossible because $f(0) < 0$ so by IVT there would be an $d; 0 < d < a$ so that $f(0)=0$. So $d in S$ but $d < inf S$ which is impossible.
And $f(a)< 0$ is impossible because $f$ is continuous, and if we let $epsilon = |f(a)| > 0$ we can find a $delta > 0$ so the for all $c; a-delta < c < a+delta$ is well be true that $|f(c)- f(a)| < epsilon$. But that means $f(a) < f(c) < 0$ so $f(c)< 0$ for all such $c$. So if $d in S$ then $d ge a+delta$ and $a + delta$ is a lower bound of $S$, contradicting that $a = inf S$.
So $f(a) = 0$ and no other value below is such.
answered Jan 28 at 3:46
fleabloodfleablood
73.4k22891
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose there is no first time. Then, for any $sin (0,1)$ such that $f(s)=0$ (why do you know that even one such $s$ exists?) there exists $s'<s$ such that $f(s')=0$ as well. We can keep doing this and get a decreasing sequence of points ${s_i}_{iinmathbb{N}}$. Does this sequence converge? How do convergent sequences behave with continuous functions?
answered Jan 28 at 3:22
Anubhav NanavatyAnubhav Nanavaty
714
$endgroup$
add a comment |
$begingroup$
Suppose there is no first time. Then, for any $sin (0,1)$ such that $f(s)=0$ (why do you know that even one such $s$ exists?) there exists $s'<s$ such that $f(s')=0$ as well. We can keep doing this and get a decreasing sequence of points ${s_i}_{iinmathbb{N}}$. Does this sequence converge? How do convergent sequences behave with continuous functions?
answered Jan 28 at 3:22
Anubhav NanavatyAnubhav Nanavaty
714
$endgroup$
add a comment |
$begingroup$
Suppose there is no first time. Then, for any $sin (0,1)$ such that $f(s)=0$ (why do you know that even one such $s$ exists?) there exists $s'<s$ such that $f(s')=0$ as well. We can keep doing this and get a decreasing sequence of points ${s_i}_{iinmathbb{N}}$. Does this sequence converge? How do convergent sequences behave with continuous functions?
answered Jan 28 at 3:22
Anubhav NanavatyAnubhav Nanavaty
714
$endgroup$
Suppose there is no first time. Then, for any $sin (0,1)$ such that $f(s)=0$ (why do you know that even one such $s$ exists?) there exists $s'<s$ such that $f(s')=0$ as well. We can keep doing this and get a decreasing sequence of points ${s_i}_{iinmathbb{N}}$. Does this sequence converge? How do convergent sequences behave with continuous functions?
answered Jan 28 at 3:22
Anubhav NanavatyAnubhav Nanavaty
714
answered Jan 28 at 3:22
Anubhav NanavatyAnubhav Nanavaty
714
answered Jan 28 at 3:22
Anubhav NanavatyAnubhav Nanavaty
714
answered Jan 28 at 3:22
Anubhav NanavatyAnubhav Nanavaty
714
714
add a comment |
add a comment |
$begingroup$
We need the following fact: Since $f$ is continuous, the set $Z(f)={xin[0,1]:f(x)=0}$ is a closed set.
It is non-empty by intermediate value theorem. Then $s=inf Z(f)=min Z(f)$ is the first time at which $f$ is zero. It is clear $sin (0,1)$.
answered Jan 28 at 3:20
Eclipse SunEclipse Sun
7,8401438
$endgroup$
add a comment |
$begingroup$
We need the following fact: Since $f$ is continuous, the set $Z(f)={xin[0,1]:f(x)=0}$ is a closed set.
It is non-empty by intermediate value theorem. Then $s=inf Z(f)=min Z(f)$ is the first time at which $f$ is zero. It is clear $sin (0,1)$.
answered Jan 28 at 3:20
Eclipse SunEclipse Sun
7,8401438
$endgroup$
add a comment |
$begingroup$
We need the following fact: Since $f$ is continuous, the set $Z(f)={xin[0,1]:f(x)=0}$ is a closed set.
It is non-empty by intermediate value theorem. Then $s=inf Z(f)=min Z(f)$ is the first time at which $f$ is zero. It is clear $sin (0,1)$.
answered Jan 28 at 3:20
Eclipse SunEclipse Sun
7,8401438
$endgroup$
We need the following fact: Since $f$ is continuous, the set $Z(f)={xin[0,1]:f(x)=0}$ is a closed set.
It is non-empty by intermediate value theorem. Then $s=inf Z(f)=min Z(f)$ is the first time at which $f$ is zero. It is clear $sin (0,1)$.
answered Jan 28 at 3:20
Eclipse SunEclipse Sun
7,8401438
answered Jan 28 at 3:20
Eclipse SunEclipse Sun
7,8401438
answered Jan 28 at 3:20
Eclipse SunEclipse Sun
7,8401438
answered Jan 28 at 3:20
Eclipse SunEclipse Sun
7,8401438
7,8401438
add a comment |
add a comment |
$begingroup$
For an argument from first principles, try defining
$$
S = {t in (0, 1) : f(x) < 0 text{ for all } x in [0, t) }.
$$
By the continuity of $f$ at $0$, $S$ is non-empty. By the continuity of $f$ at $1$, $S$ has an upper bound in $(0, 1)$. So $S$ has a least upper bound, $s in (0, 1)$. Now prove (i) $s in S$, (ii) $f(s) leqslant 0$, (iii) $f(s) geqslant 0$.
(I hope this is OK. You asked for a "hint", so I've left plenty of details to fill in. It's possible I've fooled myself!)
answered Jan 28 at 3:51
Calum GilhooleyCalum Gilhooley
5,119630
$endgroup$
add a comment |
$begingroup$
For an argument from first principles, try defining
$$
S = {t in (0, 1) : f(x) < 0 text{ for all } x in [0, t) }.
$$
By the continuity of $f$ at $0$, $S$ is non-empty. By the continuity of $f$ at $1$, $S$ has an upper bound in $(0, 1)$. So $S$ has a least upper bound, $s in (0, 1)$. Now prove (i) $s in S$, (ii) $f(s) leqslant 0$, (iii) $f(s) geqslant 0$.
(I hope this is OK. You asked for a "hint", so I've left plenty of details to fill in. It's possible I've fooled myself!)
answered Jan 28 at 3:51
Calum GilhooleyCalum Gilhooley
5,119630
$endgroup$
add a comment |
$begingroup$
For an argument from first principles, try defining
$$
S = {t in (0, 1) : f(x) < 0 text{ for all } x in [0, t) }.
$$
By the continuity of $f$ at $0$, $S$ is non-empty. By the continuity of $f$ at $1$, $S$ has an upper bound in $(0, 1)$. So $S$ has a least upper bound, $s in (0, 1)$. Now prove (i) $s in S$, (ii) $f(s) leqslant 0$, (iii) $f(s) geqslant 0$.
(I hope this is OK. You asked for a "hint", so I've left plenty of details to fill in. It's possible I've fooled myself!)
answered Jan 28 at 3:51
Calum GilhooleyCalum Gilhooley
5,119630
$endgroup$
For an argument from first principles, try defining
$$
S = {t in (0, 1) : f(x) < 0 text{ for all } x in [0, t) }.
$$
By the continuity of $f$ at $0$, $S$ is non-empty. By the continuity of $f$ at $1$, $S$ has an upper bound in $(0, 1)$. So $S$ has a least upper bound, $s in (0, 1)$. Now prove (i) $s in S$, (ii) $f(s) leqslant 0$, (iii) $f(s) geqslant 0$.
(I hope this is OK. You asked for a "hint", so I've left plenty of details to fill in. It's possible I've fooled myself!)
answered Jan 28 at 3:51
Calum GilhooleyCalum Gilhooley
5,119630
answered Jan 28 at 3:51
Calum GilhooleyCalum Gilhooley
5,119630
answered Jan 28 at 3:51
Calum GilhooleyCalum Gilhooley
5,119630
answered Jan 28 at 3:51
Calum GilhooleyCalum Gilhooley
5,119630
5,119630
add a comment |
add a comment |
$begingroup$
We know $f(c) = 0$ for some $c in [0,1]$ by the Intermediate Value Theorem.
So $S = {x|x in [0,1]; f(x) = 0}$ is not an empty set and it is bounded by $0, 1$. So $a=inf S$ exists.
What is $f(a)$.
Well $f(a) > 0$ is impossible because $f(0) < 0$ so by IVT there would be an $d; 0 < d < a$ so that $f(0)=0$. So $d in S$ but $d < inf S$ which is impossible.
And $f(a)< 0$ is impossible because $f$ is continuous, and if we let $epsilon = |f(a)| > 0$ we can find a $delta > 0$ so the for all $c; a-delta < c < a+delta$ is well be true that $|f(c)- f(a)| < epsilon$. But that means $f(a) < f(c) < 0$ so $f(c)< 0$ for all such $c$. So if $d in S$ then $d ge a+delta$ and $a + delta$ is a lower bound of $S$, contradicting that $a = inf S$.
So $f(a) = 0$ and no other value below is such.
answered Jan 28 at 3:46
fleabloodfleablood
73.4k22891
$endgroup$
add a comment |
$begingroup$
We know $f(c) = 0$ for some $c in [0,1]$ by the Intermediate Value Theorem.
So $S = {x|x in [0,1]; f(x) = 0}$ is not an empty set and it is bounded by $0, 1$. So $a=inf S$ exists.
What is $f(a)$.
Well $f(a) > 0$ is impossible because $f(0) < 0$ so by IVT there would be an $d; 0 < d < a$ so that $f(0)=0$. So $d in S$ but $d < inf S$ which is impossible.
And $f(a)< 0$ is impossible because $f$ is continuous, and if we let $epsilon = |f(a)| > 0$ we can find a $delta > 0$ so the for all $c; a-delta < c < a+delta$ is well be true that $|f(c)- f(a)| < epsilon$. But that means $f(a) < f(c) < 0$ so $f(c)< 0$ for all such $c$. So if $d in S$ then $d ge a+delta$ and $a + delta$ is a lower bound of $S$, contradicting that $a = inf S$.
So $f(a) = 0$ and no other value below is such.
answered Jan 28 at 3:46
fleabloodfleablood
73.4k22891
$endgroup$
add a comment |
$begingroup$
We know $f(c) = 0$ for some $c in [0,1]$ by the Intermediate Value Theorem.
So $S = {x|x in [0,1]; f(x) = 0}$ is not an empty set and it is bounded by $0, 1$. So $a=inf S$ exists.
What is $f(a)$.
Well $f(a) > 0$ is impossible because $f(0) < 0$ so by IVT there would be an $d; 0 < d < a$ so that $f(0)=0$. So $d in S$ but $d < inf S$ which is impossible.
And $f(a)< 0$ is impossible because $f$ is continuous, and if we let $epsilon = |f(a)| > 0$ we can find a $delta > 0$ so the for all $c; a-delta < c < a+delta$ is well be true that $|f(c)- f(a)| < epsilon$. But that means $f(a) < f(c) < 0$ so $f(c)< 0$ for all such $c$. So if $d in S$ then $d ge a+delta$ and $a + delta$ is a lower bound of $S$, contradicting that $a = inf S$.
So $f(a) = 0$ and no other value below is such.
answered Jan 28 at 3:46
fleabloodfleablood
73.4k22891
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We know $f(c) = 0$ for some $c in [0,1]$ by the Intermediate Value Theorem.
So $S = {x|x in [0,1]; f(x) = 0}$ is not an empty set and it is bounded by $0, 1$. So $a=inf S$ exists.
What is $f(a)$.
Well $f(a) > 0$ is impossible because $f(0) < 0$ so by IVT there would be an $d; 0 < d < a$ so that $f(0)=0$. So $d in S$ but $d < inf S$ which is impossible.
And $f(a)< 0$ is impossible because $f$ is continuous, and if we let $epsilon = |f(a)| > 0$ we can find a $delta > 0$ so the for all $c; a-delta < c < a+delta$ is well be true that $|f(c)- f(a)| < epsilon$. But that means $f(a) < f(c) < 0$ so $f(c)< 0$ for all such $c$. So if $d in S$ then $d ge a+delta$ and $a + delta$ is a lower bound of $S$, contradicting that $a = inf S$.
So $f(a) = 0$ and no other value below is such.
answered Jan 28 at 3:46
fleabloodfleablood
73.4k22891
answered Jan 28 at 3:46
fleabloodfleablood
73.4k22891
answered Jan 28 at 3:46
fleabloodfleablood
73.4k22891
answered Jan 28 at 3:46
fleabloodfleablood
73.4k22891
73.4k22891
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$begingroup$
You should use the supremum to come up with a candidate for "the first time at which $f$ is zero". Can you think of such a candidate?
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– астон вілла олоф мэллбэрг
Jan 28 at 3:20
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Intermediate Value Theorem tells us that there is at least one root. Now, can we prove that given this case, the function has at most countably many roots?
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– Aniruddha Deshmukh
Jan 28 at 3:20
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@AniruddhaDeshmukh no-one said anything about countably many roots and there's no reason to assume countablity. Consider $f(x) = 0$ for all $frac 14 le x le frac 34$ but $f(x) < 0$ for $x < frac 14$ and $f(x) > 0$ for $x >frac 34$. That has uncountably many roots but $x = frac 14$ is the least ("first") $x$ so that $f(x) = 0$.
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– fleablood
Jan 28 at 3:35
1
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Somebody asked exactly the same question few hours ago: math.stackexchange.com/questions/3090223/…
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– N. S.
Jan 28 at 3:48
1
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Possible duplicate of A function is continuous. Let $f : [0, 1] to mathbb R$ satisfy $f(0) = −1$, $f(1) = 1$. Prove that there is a first time at which f is zero
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– N. S.
Jan 28 at 3:48