Prove the support of a real function is countable
Clash Royale CLAN TAG#URR8PPP
$begingroup$
This is a problem from my real analysis homework. We are learning the countable sets, and have yet to reach uncountable sets.
Let $f$ be a real function defined on $[0, 1]$. There exists a constant $M$, such that for each finite $n$, and $0 le x_1 < x_2 < cdots < x_n le 1$, we have
$$
|f(x_1) + f(x_2) + cdots + f(x_n)| le M.
$$
Prove $E stackreltextdef= x in [0, 1] : f(x) ne 0$ is countable.
My attempt
I split $E$ into two parts,
$$
beginalign
E^+ &stackreltextdef= x in [0, 1] : f(x) > 0 \
E^- &stackreltextdef= x in [0, 1] : f(x) < 0
endalign
$$
A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, but I didn't gain much from her hint.
Another path I have taken is letting
$$
beginalign
E' &stackreltextdef= x in E : exists delta_x > 0, (x, x+delta_x) cap E = varnothing \
E'' &stackreltextdef= E setminus E'
endalign
$$
I can prove $E'$ is countable, but $E''$ is still tricky even though it contains less or equal elements than $E$.
Proof by contradiction didn't yield any significant result, either.
real-analysis analysis
edited Mar 2 at 8:09
Asaf Karagila♦
307k33439771
asked Mar 2 at 8:04
nalzoknalzok
291217
$endgroup$
|
show 3 more comments
$begingroup$
This is a problem from my real analysis homework. We are learning the countable sets, and have yet to reach uncountable sets.
Let $f$ be a real function defined on $[0, 1]$. There exists a constant $M$, such that for each finite $n$, and $0 le x_1 < x_2 < cdots < x_n le 1$, we have
$$
|f(x_1) + f(x_2) + cdots + f(x_n)| le M.
$$
Prove $E stackreltextdef= x in [0, 1] : f(x) ne 0$ is countable.
My attempt
I split $E$ into two parts,
$$
beginalign
E^+ &stackreltextdef= x in [0, 1] : f(x) > 0 \
E^- &stackreltextdef= x in [0, 1] : f(x) < 0
endalign
$$
A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, but I didn't gain much from her hint.
Another path I have taken is letting
$$
beginalign
E' &stackreltextdef= x in E : exists delta_x > 0, (x, x+delta_x) cap E = varnothing \
E'' &stackreltextdef= E setminus E'
endalign
$$
I can prove $E'$ is countable, but $E''$ is still tricky even though it contains less or equal elements than $E$.
Proof by contradiction didn't yield any significant result, either.
real-analysis analysis
edited Mar 2 at 8:09
Asaf Karagila♦
307k33439771
asked Mar 2 at 8:04
nalzoknalzok
291217
$endgroup$
3
$begingroup$
Terminological remark: usually, the set $E$ you define is not called the support, but rather its closure is. For instance, if you consider a function which is nonzero precisely at the rational numbers, then its support would be all of $[0,1]$. This doesn't impact the question because of how it's phrased, but it's something to be aware of.
$endgroup$
– Wojowu
Mar 2 at 9:20
1
$begingroup$
The trick is always the same; the nonzero reals can be partitioned into countably many sets each of which is strictly bounded away from zero. Remember this trick for real analysis and measure theory.
$endgroup$
– user21820
Mar 2 at 10:06
$begingroup$
@Wojowu Thanks for pointing that out! Initially I wasn't sure if it's the correct term, so I consulted Wikipedia, and that appears to be the definition of a support? To quote, "In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero."
$endgroup$
– nalzok
Mar 2 at 10:52
$begingroup$
Your classmate's suggestion might be slightly better as considering $f(x) > frac1n$ and $f(x) lt -frac1n$ from $E$ separately, and proving for each $n$ both parts have a finite number of elements
$endgroup$
– Henry
Mar 2 at 13:50
$begingroup$
@Henry I could have misunderstood her point...
$endgroup$
– nalzok
Mar 2 at 13:56
|
show 3 more comments
$begingroup$
This is a problem from my real analysis homework. We are learning the countable sets, and have yet to reach uncountable sets.
Let $f$ be a real function defined on $[0, 1]$. There exists a constant $M$, such that for each finite $n$, and $0 le x_1 < x_2 < cdots < x_n le 1$, we have
$$
|f(x_1) + f(x_2) + cdots + f(x_n)| le M.
$$
Prove $E stackreltextdef= x in [0, 1] : f(x) ne 0$ is countable.
My attempt
I split $E$ into two parts,
$$
beginalign
E^+ &stackreltextdef= x in [0, 1] : f(x) > 0 \
E^- &stackreltextdef= x in [0, 1] : f(x) < 0
endalign
$$
A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, but I didn't gain much from her hint.
Another path I have taken is letting
$$
beginalign
E' &stackreltextdef= x in E : exists delta_x > 0, (x, x+delta_x) cap E = varnothing \
E'' &stackreltextdef= E setminus E'
endalign
$$
I can prove $E'$ is countable, but $E''$ is still tricky even though it contains less or equal elements than $E$.
Proof by contradiction didn't yield any significant result, either.
real-analysis analysis
edited Mar 2 at 8:09
Asaf Karagila♦
307k33439771
asked Mar 2 at 8:04
nalzoknalzok
291217
$endgroup$
This is a problem from my real analysis homework. We are learning the countable sets, and have yet to reach uncountable sets.
Let $f$ be a real function defined on $[0, 1]$. There exists a constant $M$, such that for each finite $n$, and $0 le x_1 < x_2 < cdots < x_n le 1$, we have
$$
|f(x_1) + f(x_2) + cdots + f(x_n)| le M.
$$
Prove $E stackreltextdef= x in [0, 1] : f(x) ne 0$ is countable.
My attempt
I split $E$ into two parts,
$$
beginalign
E^+ &stackreltextdef= x in [0, 1] : f(x) > 0 \
E^- &stackreltextdef= x in [0, 1] : f(x) < 0
endalign
$$
A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, but I didn't gain much from her hint.
Another path I have taken is letting
$$
beginalign
E' &stackreltextdef= x in E : exists delta_x > 0, (x, x+delta_x) cap E = varnothing \
E'' &stackreltextdef= E setminus E'
endalign
$$
I can prove $E'$ is countable, but $E''$ is still tricky even though it contains less or equal elements than $E$.
Proof by contradiction didn't yield any significant result, either.
real-analysis analysis
real-analysis analysis
edited Mar 2 at 8:09
Asaf Karagila♦
307k33439771
asked Mar 2 at 8:04
nalzoknalzok
291217
edited Mar 2 at 8:09
Asaf Karagila♦
307k33439771
asked Mar 2 at 8:04
nalzoknalzok
291217
edited Mar 2 at 8:09
Asaf Karagila♦
307k33439771
edited Mar 2 at 8:09
Asaf Karagila♦
307k33439771
edited Mar 2 at 8:09
Asaf Karagila♦
307k33439771
307k33439771
asked Mar 2 at 8:04
nalzoknalzok
291217
asked Mar 2 at 8:04
nalzoknalzok
291217
asked Mar 2 at 8:04
nalzoknalzok
291217
291217
3
$begingroup$
Terminological remark: usually, the set $E$ you define is not called the support, but rather its closure is. For instance, if you consider a function which is nonzero precisely at the rational numbers, then its support would be all of $[0,1]$. This doesn't impact the question because of how it's phrased, but it's something to be aware of.
$endgroup$
– Wojowu
Mar 2 at 9:20
1
$begingroup$
The trick is always the same; the nonzero reals can be partitioned into countably many sets each of which is strictly bounded away from zero. Remember this trick for real analysis and measure theory.
$endgroup$
– user21820
Mar 2 at 10:06
$begingroup$
@Wojowu Thanks for pointing that out! Initially I wasn't sure if it's the correct term, so I consulted Wikipedia, and that appears to be the definition of a support? To quote, "In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero."
$endgroup$
– nalzok
Mar 2 at 10:52
$begingroup$
Your classmate's suggestion might be slightly better as considering $f(x) > frac1n$ and $f(x) lt -frac1n$ from $E$ separately, and proving for each $n$ both parts have a finite number of elements
$endgroup$
– Henry
Mar 2 at 13:50
$begingroup$
@Henry I could have misunderstood her point...
$endgroup$
– nalzok
Mar 2 at 13:56
|
show 3 more comments
3
$begingroup$
Terminological remark: usually, the set $E$ you define is not called the support, but rather its closure is. For instance, if you consider a function which is nonzero precisely at the rational numbers, then its support would be all of $[0,1]$. This doesn't impact the question because of how it's phrased, but it's something to be aware of.
$endgroup$
– Wojowu
Mar 2 at 9:20
1
$begingroup$
The trick is always the same; the nonzero reals can be partitioned into countably many sets each of which is strictly bounded away from zero. Remember this trick for real analysis and measure theory.
$endgroup$
– user21820
Mar 2 at 10:06
$begingroup$
@Wojowu Thanks for pointing that out! Initially I wasn't sure if it's the correct term, so I consulted Wikipedia, and that appears to be the definition of a support? To quote, "In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero."
$endgroup$
– nalzok
Mar 2 at 10:52
$begingroup$
Your classmate's suggestion might be slightly better as considering $f(x) > frac1n$ and $f(x) lt -frac1n$ from $E$ separately, and proving for each $n$ both parts have a finite number of elements
$endgroup$
– Henry
Mar 2 at 13:50
$begingroup$
@Henry I could have misunderstood her point...
$endgroup$
– nalzok
Mar 2 at 13:56
3
3
$begingroup$
Terminological remark: usually, the set $E$ you define is not called the support, but rather its closure is. For instance, if you consider a function which is nonzero precisely at the rational numbers, then its support would be all of $[0,1]$. This doesn't impact the question because of how it's phrased, but it's something to be aware of.
$endgroup$
– Wojowu
Mar 2 at 9:20
$begingroup$
Terminological remark: usually, the set $E$ you define is not called the support, but rather its closure is. For instance, if you consider a function which is nonzero precisely at the rational numbers, then its support would be all of $[0,1]$. This doesn't impact the question because of how it's phrased, but it's something to be aware of.
$endgroup$
– Wojowu
Mar 2 at 9:20
1
1
$begingroup$
The trick is always the same; the nonzero reals can be partitioned into countably many sets each of which is strictly bounded away from zero. Remember this trick for real analysis and measure theory.
$endgroup$
– user21820
Mar 2 at 10:06
$begingroup$
The trick is always the same; the nonzero reals can be partitioned into countably many sets each of which is strictly bounded away from zero. Remember this trick for real analysis and measure theory.
$endgroup$
– user21820
Mar 2 at 10:06
$begingroup$
@Wojowu Thanks for pointing that out! Initially I wasn't sure if it's the correct term, so I consulted Wikipedia, and that appears to be the definition of a support? To quote, "In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero."
$endgroup$
– nalzok
Mar 2 at 10:52
$begingroup$
@Wojowu Thanks for pointing that out! Initially I wasn't sure if it's the correct term, so I consulted Wikipedia, and that appears to be the definition of a support? To quote, "In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero."
$endgroup$
– nalzok
Mar 2 at 10:52
$begingroup$
Your classmate's suggestion might be slightly better as considering $f(x) > frac1n$ and $f(x) lt -frac1n$ from $E$ separately, and proving for each $n$ both parts have a finite number of elements
$endgroup$
– Henry
Mar 2 at 13:50
$begingroup$
Your classmate's suggestion might be slightly better as considering $f(x) > frac1n$ and $f(x) lt -frac1n$ from $E$ separately, and proving for each $n$ both parts have a finite number of elements
$endgroup$
– Henry
Mar 2 at 13:50
$begingroup$
@Henry I could have misunderstood her point...
$endgroup$
– nalzok
Mar 2 at 13:56
$begingroup$
@Henry I could have misunderstood her point...
$endgroup$
– nalzok
Mar 2 at 13:56
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, ...
Your classmate is on the right track, but it suffices to show that there are (at most) finitely many points with $f(x) > frac1n$ in $E^+$.
For each $n in Bbb N = 1, 2, 3, ldots $ define
$$
E^+_n = leftx in [0, 1] : f(x) > frac 1n right
$$
Then $E^+_n$ has at most $nM$ elements. It follows that
$$
E^+ = bigcup_n in Bbb N E^+_n = x in [0, 1] : f(x) > 0
$$
is countable (as the countable union of finite sets).
In the same way (or by applying the above argument to $-f$) it can be shown that $E^-$ is countable as well.
Remark: The domain of $f$ (in your case: the interval $[0, 1]$) is irrelevant for this conclusion. The same statement holds for a real-valued function $f$ defined on an arbitrary set $X$.
answered Mar 2 at 8:10
Martin RMartin R
30.5k33558
$endgroup$
add a comment |
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$begingroup$
A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, ...
Your classmate is on the right track, but it suffices to show that there are (at most) finitely many points with $f(x) > frac1n$ in $E^+$.
For each $n in Bbb N = 1, 2, 3, ldots $ define
$$
E^+_n = leftx in [0, 1] : f(x) > frac 1n right
$$
Then $E^+_n$ has at most $nM$ elements. It follows that
$$
E^+ = bigcup_n in Bbb N E^+_n = x in [0, 1] : f(x) > 0
$$
is countable (as the countable union of finite sets).
In the same way (or by applying the above argument to $-f$) it can be shown that $E^-$ is countable as well.
Remark: The domain of $f$ (in your case: the interval $[0, 1]$) is irrelevant for this conclusion. The same statement holds for a real-valued function $f$ defined on an arbitrary set $X$.
answered Mar 2 at 8:10
Martin RMartin R
30.5k33558
$endgroup$
add a comment |
$begingroup$
A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, ...
Your classmate is on the right track, but it suffices to show that there are (at most) finitely many points with $f(x) > frac1n$ in $E^+$.
For each $n in Bbb N = 1, 2, 3, ldots $ define
$$
E^+_n = leftx in [0, 1] : f(x) > frac 1n right
$$
Then $E^+_n$ has at most $nM$ elements. It follows that
$$
E^+ = bigcup_n in Bbb N E^+_n = x in [0, 1] : f(x) > 0
$$
is countable (as the countable union of finite sets).
In the same way (or by applying the above argument to $-f$) it can be shown that $E^-$ is countable as well.
Remark: The domain of $f$ (in your case: the interval $[0, 1]$) is irrelevant for this conclusion. The same statement holds for a real-valued function $f$ defined on an arbitrary set $X$.
answered Mar 2 at 8:10
Martin RMartin R
30.5k33558
$endgroup$
add a comment |
$begingroup$
A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, ...
Your classmate is on the right track, but it suffices to show that there are (at most) finitely many points with $f(x) > frac1n$ in $E^+$.
For each $n in Bbb N = 1, 2, 3, ldots $ define
$$
E^+_n = leftx in [0, 1] : f(x) > frac 1n right
$$
Then $E^+_n$ has at most $nM$ elements. It follows that
$$
E^+ = bigcup_n in Bbb N E^+_n = x in [0, 1] : f(x) > 0
$$
is countable (as the countable union of finite sets).
In the same way (or by applying the above argument to $-f$) it can be shown that $E^-$ is countable as well.
Remark: The domain of $f$ (in your case: the interval $[0, 1]$) is irrelevant for this conclusion. The same statement holds for a real-valued function $f$ defined on an arbitrary set $X$.
answered Mar 2 at 8:10
Martin RMartin R
30.5k33558
$endgroup$
A classmate suggested considering $f(x) > frac1n$ and $f(x) le frac1n$ from $E^+$ separately, and proving for each $n$ both part have finite amount of elements, ...
Your classmate is on the right track, but it suffices to show that there are (at most) finitely many points with $f(x) > frac1n$ in $E^+$.
For each $n in Bbb N = 1, 2, 3, ldots $ define
$$
E^+_n = leftx in [0, 1] : f(x) > frac 1n right
$$
Then $E^+_n$ has at most $nM$ elements. It follows that
$$
E^+ = bigcup_n in Bbb N E^+_n = x in [0, 1] : f(x) > 0
$$
is countable (as the countable union of finite sets).
In the same way (or by applying the above argument to $-f$) it can be shown that $E^-$ is countable as well.
Remark: The domain of $f$ (in your case: the interval $[0, 1]$) is irrelevant for this conclusion. The same statement holds for a real-valued function $f$ defined on an arbitrary set $X$.
answered Mar 2 at 8:10
Martin RMartin R
30.5k33558
edited Mar 2 at 15:37
answered Mar 2 at 8:10
Martin RMartin R
30.5k33558
answered Mar 2 at 8:10
Martin RMartin R
30.5k33558
answered Mar 2 at 8:10
Martin RMartin R
30.5k33558
30.5k33558
add a comment |
add a comment |
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3
$begingroup$
Terminological remark: usually, the set $E$ you define is not called the support, but rather its closure is. For instance, if you consider a function which is nonzero precisely at the rational numbers, then its support would be all of $[0,1]$. This doesn't impact the question because of how it's phrased, but it's something to be aware of.
$endgroup$
– Wojowu
Mar 2 at 9:20
1
$begingroup$
The trick is always the same; the nonzero reals can be partitioned into countably many sets each of which is strictly bounded away from zero. Remember this trick for real analysis and measure theory.
$endgroup$
– user21820
Mar 2 at 10:06
$begingroup$
@Wojowu Thanks for pointing that out! Initially I wasn't sure if it's the correct term, so I consulted Wikipedia, and that appears to be the definition of a support? To quote, "In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero."
$endgroup$
– nalzok
Mar 2 at 10:52
$begingroup$
Your classmate's suggestion might be slightly better as considering $f(x) > frac1n$ and $f(x) lt -frac1n$ from $E$ separately, and proving for each $n$ both parts have a finite number of elements
$endgroup$
– Henry
Mar 2 at 13:50
$begingroup$
@Henry I could have misunderstood her point...
$endgroup$
– nalzok
Mar 2 at 13:56