Prove the support of a real function is countable
$begingroup$
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 $E stackrel{text{def}}{=} {x in [0, 1] : f(x) ne 0}$ is countable.
My attempt
I split $E$ into two parts,
$$
begin{align}
E^{+} &stackrel{text{def}}{=} {x in [0, 1] : f(x) > 0} \
E^{-} &stackrel{text{def}}{=} {x in [0, 1] : f(x) < 0}
end{align}
$$
A classmate suggested considering $f(x) > frac{1}{n}$ and $f(x) le frac{1}{n}$ 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
$$
begin{align}
E' &stackrel{text{def}}{=} {x in E : exists delta_x > 0, (x, x+delta_x) cap E = varnothing } \
E'' &stackrel{text{def}}{=} E setminus E'
end{align}
$$
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♦
307k33441774
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 $E stackrel{text{def}}{=} {x in [0, 1] : f(x) ne 0}$ is countable.
My attempt
I split $E$ into two parts,
$$
begin{align}
E^{+} &stackrel{text{def}}{=} {x in [0, 1] : f(x) > 0} \
E^{-} &stackrel{text{def}}{=} {x in [0, 1] : f(x) < 0}
end{align}
$$
A classmate suggested considering $f(x) > frac{1}{n}$ and $f(x) le frac{1}{n}$ 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
$$
begin{align}
E' &stackrel{text{def}}{=} {x in E : exists delta_x > 0, (x, x+delta_x) cap E = varnothing } \
E'' &stackrel{text{def}}{=} E setminus E'
end{align}
$$
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♦
307k33441774
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 $f(x) > frac{1}{n}$ and $f(x) lt -frac{1}{n}$ 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 $E stackrel{text{def}}{=} {x in [0, 1] : f(x) ne 0}$ is countable.
My attempt
I split $E$ into two parts,
$$
begin{align}
E^{+} &stackrel{text{def}}{=} {x in [0, 1] : f(x) > 0} \
E^{-} &stackrel{text{def}}{=} {x in [0, 1] : f(x) < 0}
end{align}
$$
A classmate suggested considering $f(x) > frac{1}{n}$ and $f(x) le frac{1}{n}$ 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
$$
begin{align}
E' &stackrel{text{def}}{=} {x in E : exists delta_x > 0, (x, x+delta_x) cap E = varnothing } \
E'' &stackrel{text{def}}{=} E setminus E'
end{align}
$$
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♦
307k33441774
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 $E stackrel{text{def}}{=} {x in [0, 1] : f(x) ne 0}$ is countable.
My attempt
I split $E$ into two parts,
$$
begin{align}
E^{+} &stackrel{text{def}}{=} {x in [0, 1] : f(x) > 0} \
E^{-} &stackrel{text{def}}{=} {x in [0, 1] : f(x) < 0}
end{align}
$$
A classmate suggested considering $f(x) > frac{1}{n}$ and $f(x) le frac{1}{n}$ 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
$$
begin{align}
E' &stackrel{text{def}}{=} {x in E : exists delta_x > 0, (x, x+delta_x) cap E = varnothing } \
E'' &stackrel{text{def}}{=} E setminus E'
end{align}
$$
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♦
307k33441774
asked Mar 2 at 8:04
nalzoknalzok
291217
edited Mar 2 at 8:09
Asaf Karagila♦
307k33441774
asked Mar 2 at 8:04
nalzoknalzok
291217
edited Mar 2 at 8:09
Asaf Karagila♦
307k33441774
edited Mar 2 at 8:09
Asaf Karagila♦
307k33441774
edited Mar 2 at 8:09
Asaf Karagila♦
307k33441774
307k33441774
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 $f(x) > frac{1}{n}$ and $f(x) lt -frac{1}{n}$ 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 $f(x) > frac{1}{n}$ and $f(x) lt -frac{1}{n}$ 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 $f(x) > frac{1}{n}$ and $f(x) lt -frac{1}{n}$ 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 $f(x) > frac{1}{n}$ and $f(x) lt -frac{1}{n}$ 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 $f(x) > frac{1}{n}$ and $f(x) le frac{1}{n}$ 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 $f(x) > frac{1}{n}$ in $E^{+}$.
For each $n in Bbb N = { 1, 2, 3, ldots }$ define
$$
E^{+}_n = left{x in [0, 1] : f(x) > frac 1n right}
$$
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
30.7k33560
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3132179%2fprove-the-support-of-a-real-function-is-countable%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A classmate suggested considering $f(x) > frac{1}{n}$ and $f(x) le frac{1}{n}$ 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 $f(x) > frac{1}{n}$ in $E^{+}$.
For each $n in Bbb N = { 1, 2, 3, ldots }$ define
$$
E^{+}_n = left{x in [0, 1] : f(x) > frac 1n right}
$$
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
30.7k33560
$endgroup$
add a comment |
$begingroup$
A classmate suggested considering $f(x) > frac{1}{n}$ and $f(x) le frac{1}{n}$ 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 $f(x) > frac{1}{n}$ in $E^{+}$.
For each $n in Bbb N = { 1, 2, 3, ldots }$ define
$$
E^{+}_n = left{x in [0, 1] : f(x) > frac 1n right}
$$
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
30.7k33560
$endgroup$
add a comment |
$begingroup$
A classmate suggested considering $f(x) > frac{1}{n}$ and $f(x) le frac{1}{n}$ 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 $f(x) > frac{1}{n}$ in $E^{+}$.
For each $n in Bbb N = { 1, 2, 3, ldots }$ define
$$
E^{+}_n = left{x in [0, 1] : f(x) > frac 1n right}
$$
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
30.7k33560
$endgroup$
A classmate suggested considering $f(x) > frac{1}{n}$ and $f(x) le frac{1}{n}$ 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 $f(x) > frac{1}{n}$ in $E^{+}$.
For each $n in Bbb N = { 1, 2, 3, ldots }$ define
$$
E^{+}_n = left{x in [0, 1] : f(x) > frac 1n right}
$$
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
30.7k33560
edited Mar 2 at 15:37
answered Mar 2 at 8:10
Martin RMartin R
30.7k33560
answered Mar 2 at 8:10
Martin RMartin R
30.7k33560
answered Mar 2 at 8:10
Martin RMartin R
30.7k33560
30.7k33560
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3132179%2fprove-the-support-of-a-real-function-is-countable%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 $f(x) > frac{1}{n}$ and $f(x) lt -frac{1}{n}$ 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