Mixing
Showing that the Dirichlet function has no limit
Let $f(x) = 0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational. Prove that $lim_{xto a} f(x)$ does not exist for any $a$.
Proof attempt:
Let $ainmathbb{R}$.
The only two possible values for a limit of $lim_{xto a}f(x)$ are $0,1$, since otherwise there will always be a constant gap between the valules of $f$ and the limit, and hence $f$ would not get arbitrarily small.
However, $lim_{xto a}f(x) not = 1$ for the following reason: Set $epsilon = 1/2$ and let $delta > 0$. Then there exists a rational number $x_0$ in $0<|x_0-a|<delta$ by the density of the rationals. However, $|f(x_0)-1| = |0-1| = 1 not < 1/2$, and so $lim_{xto a}f(x) not = 1$.
Also, $lim_{xto a}f(x) not = 0$ for the following reason: Set $epsilon = 1/2$ and let $delta > 0$. Then there exists a irrational number $x_0$ in $0<|x_0-a|<delta$ by the density of the irrationals. However, $|f(x_0)-0| = |1-0| = 1 not < 1/2$, and so $lim_{xto a}f(x) not = 0$.
real-analysis limits proof-verification
asked Nov 21 '18 at 4:11
Wesley
525313
add a comment |
Let $f(x) = 0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational. Prove that $lim_{xto a} f(x)$ does not exist for any $a$.
Proof attempt:
Let $ainmathbb{R}$.
The only two possible values for a limit of $lim_{xto a}f(x)$ are $0,1$, since otherwise there will always be a constant gap between the valules of $f$ and the limit, and hence $f$ would not get arbitrarily small.
However, $lim_{xto a}f(x) not = 1$ for the following reason: Set $epsilon = 1/2$ and let $delta > 0$. Then there exists a rational number $x_0$ in $0<|x_0-a|<delta$ by the density of the rationals. However, $|f(x_0)-1| = |0-1| = 1 not < 1/2$, and so $lim_{xto a}f(x) not = 1$.
Also, $lim_{xto a}f(x) not = 0$ for the following reason: Set $epsilon = 1/2$ and let $delta > 0$. Then there exists a irrational number $x_0$ in $0<|x_0-a|<delta$ by the density of the irrationals. However, $|f(x_0)-0| = |1-0| = 1 not < 1/2$, and so $lim_{xto a}f(x) not = 0$.
real-analysis limits proof-verification
asked Nov 21 '18 at 4:11
Wesley
525313
Exactly what are you asking? What point is the one you are unsure about?
– Will M.
Nov 21 '18 at 4:13
I am unsure of whether my proof is correct. For example, is assuming that the only possible limits are 0 and 1 a correct assumption? Then do I correctly rule these out as limits?
– Wesley
Nov 21 '18 at 4:15
Ok, technically speaking, you need to consider a value, whatever it may be, $L in mathbf{R},$ and then rule it out as a possible limit. You can divide three cases: (1) $L neq 0, 1$ (2) $L = 0$ and (3) $L=1.$
– Will M.
Nov 21 '18 at 4:17
Is at least my reasoning for cases (2) and (3) correct?
– Wesley
Nov 21 '18 at 4:30
add a comment |
Let $f(x) = 0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational. Prove that $lim_{xto a} f(x)$ does not exist for any $a$.
Proof attempt:
Let $ainmathbb{R}$.
The only two possible values for a limit of $lim_{xto a}f(x)$ are $0,1$, since otherwise there will always be a constant gap between the valules of $f$ and the limit, and hence $f$ would not get arbitrarily small.
However, $lim_{xto a}f(x) not = 1$ for the following reason: Set $epsilon = 1/2$ and let $delta > 0$. Then there exists a rational number $x_0$ in $0<|x_0-a|<delta$ by the density of the rationals. However, $|f(x_0)-1| = |0-1| = 1 not < 1/2$, and so $lim_{xto a}f(x) not = 1$.
Also, $lim_{xto a}f(x) not = 0$ for the following reason: Set $epsilon = 1/2$ and let $delta > 0$. Then there exists a irrational number $x_0$ in $0<|x_0-a|<delta$ by the density of the irrationals. However, $|f(x_0)-0| = |1-0| = 1 not < 1/2$, and so $lim_{xto a}f(x) not = 0$.
real-analysis limits proof-verification
asked Nov 21 '18 at 4:11
Wesley
525313
Let $f(x) = 0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational. Prove that $lim_{xto a} f(x)$ does not exist for any $a$.
Proof attempt:
Let $ainmathbb{R}$.
The only two possible values for a limit of $lim_{xto a}f(x)$ are $0,1$, since otherwise there will always be a constant gap between the valules of $f$ and the limit, and hence $f$ would not get arbitrarily small.
However, $lim_{xto a}f(x) not = 1$ for the following reason: Set $epsilon = 1/2$ and let $delta > 0$. Then there exists a rational number $x_0$ in $0<|x_0-a|<delta$ by the density of the rationals. However, $|f(x_0)-1| = |0-1| = 1 not < 1/2$, and so $lim_{xto a}f(x) not = 1$.
Also, $lim_{xto a}f(x) not = 0$ for the following reason: Set $epsilon = 1/2$ and let $delta > 0$. Then there exists a irrational number $x_0$ in $0<|x_0-a|<delta$ by the density of the irrationals. However, $|f(x_0)-0| = |1-0| = 1 not < 1/2$, and so $lim_{xto a}f(x) not = 0$.
real-analysis limits proof-verification
real-analysis limits proof-verification
asked Nov 21 '18 at 4:11
Wesley
525313
asked Nov 21 '18 at 4:11
Wesley
525313
asked Nov 21 '18 at 4:11
Wesley
525313
asked Nov 21 '18 at 4:11
Wesley
525313
asked Nov 21 '18 at 4:11
Wesley
525313
525313
Exactly what are you asking? What point is the one you are unsure about?
– Will M.
Nov 21 '18 at 4:13
I am unsure of whether my proof is correct. For example, is assuming that the only possible limits are 0 and 1 a correct assumption? Then do I correctly rule these out as limits?
– Wesley
Nov 21 '18 at 4:15
Ok, technically speaking, you need to consider a value, whatever it may be, $L in mathbf{R},$ and then rule it out as a possible limit. You can divide three cases: (1) $L neq 0, 1$ (2) $L = 0$ and (3) $L=1.$
– Will M.
Nov 21 '18 at 4:17
Is at least my reasoning for cases (2) and (3) correct?
– Wesley
Nov 21 '18 at 4:30
add a comment |
Exactly what are you asking? What point is the one you are unsure about?
– Will M.
Nov 21 '18 at 4:13
I am unsure of whether my proof is correct. For example, is assuming that the only possible limits are 0 and 1 a correct assumption? Then do I correctly rule these out as limits?
– Wesley
Nov 21 '18 at 4:15
Ok, technically speaking, you need to consider a value, whatever it may be, $L in mathbf{R},$ and then rule it out as a possible limit. You can divide three cases: (1) $L neq 0, 1$ (2) $L = 0$ and (3) $L=1.$
– Will M.
Nov 21 '18 at 4:17
Is at least my reasoning for cases (2) and (3) correct?
– Wesley
Nov 21 '18 at 4:30
Exactly what are you asking? What point is the one you are unsure about?
– Will M.
Nov 21 '18 at 4:13
Exactly what are you asking? What point is the one you are unsure about?
– Will M.
Nov 21 '18 at 4:13
I am unsure of whether my proof is correct. For example, is assuming that the only possible limits are 0 and 1 a correct assumption? Then do I correctly rule these out as limits?
– Wesley
Nov 21 '18 at 4:15
I am unsure of whether my proof is correct. For example, is assuming that the only possible limits are 0 and 1 a correct assumption? Then do I correctly rule these out as limits?
– Wesley
Nov 21 '18 at 4:15
Ok, technically speaking, you need to consider a value, whatever it may be, $L in mathbf{R},$ and then rule it out as a possible limit. You can divide three cases: (1) $L neq 0, 1$ (2) $L = 0$ and (3) $L=1.$
– Will M.
Nov 21 '18 at 4:17
Ok, technically speaking, you need to consider a value, whatever it may be, $L in mathbf{R},$ and then rule it out as a possible limit. You can divide three cases: (1) $L neq 0, 1$ (2) $L = 0$ and (3) $L=1.$
– Will M.
Nov 21 '18 at 4:17
Is at least my reasoning for cases (2) and (3) correct?
– Wesley
Nov 21 '18 at 4:30
Is at least my reasoning for cases (2) and (3) correct?
– Wesley
Nov 21 '18 at 4:30
add a comment |
1 Answer
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votes
Your proof is complete and fine. Only one minor correction : "hence $f$ would not get arbitrarily small" should be changed to "hence $f$ would not get arbitrarily close to the possible limit", in the first paragraph of the proof (or changed to "hence $f-L$ would not get arbitrarily small, where $L neq 0,1$ is a possible limit").
There is another proof that will avoid the breaking of $L$ into cases : indeed, given $a in mathbb R$, by density of rationals and irrationals, there is a sequence of rationals and irrationals converging to $a$. But, $lim_{x to a} f(x)$ exists if and only if for any two sequences $a_n, b_n to a$ we have $lim a_n = lim b_n$, and here the sequence of rationals and irrationals produce different values, namely $0$ and $1$.
answered Nov 21 '18 at 5:02
астон вілла олоф мэллбэрг
37.3k33376
add a comment |
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1 Answer
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Your proof is complete and fine. Only one minor correction : "hence $f$ would not get arbitrarily small" should be changed to "hence $f$ would not get arbitrarily close to the possible limit", in the first paragraph of the proof (or changed to "hence $f-L$ would not get arbitrarily small, where $L neq 0,1$ is a possible limit").
There is another proof that will avoid the breaking of $L$ into cases : indeed, given $a in mathbb R$, by density of rationals and irrationals, there is a sequence of rationals and irrationals converging to $a$. But, $lim_{x to a} f(x)$ exists if and only if for any two sequences $a_n, b_n to a$ we have $lim a_n = lim b_n$, and here the sequence of rationals and irrationals produce different values, namely $0$ and $1$.
answered Nov 21 '18 at 5:02
астон вілла олоф мэллбэрг
37.3k33376
add a comment |
Your proof is complete and fine. Only one minor correction : "hence $f$ would not get arbitrarily small" should be changed to "hence $f$ would not get arbitrarily close to the possible limit", in the first paragraph of the proof (or changed to "hence $f-L$ would not get arbitrarily small, where $L neq 0,1$ is a possible limit").
There is another proof that will avoid the breaking of $L$ into cases : indeed, given $a in mathbb R$, by density of rationals and irrationals, there is a sequence of rationals and irrationals converging to $a$. But, $lim_{x to a} f(x)$ exists if and only if for any two sequences $a_n, b_n to a$ we have $lim a_n = lim b_n$, and here the sequence of rationals and irrationals produce different values, namely $0$ and $1$.
answered Nov 21 '18 at 5:02
астон вілла олоф мэллбэрг
37.3k33376
add a comment |
Your proof is complete and fine. Only one minor correction : "hence $f$ would not get arbitrarily small" should be changed to "hence $f$ would not get arbitrarily close to the possible limit", in the first paragraph of the proof (or changed to "hence $f-L$ would not get arbitrarily small, where $L neq 0,1$ is a possible limit").
There is another proof that will avoid the breaking of $L$ into cases : indeed, given $a in mathbb R$, by density of rationals and irrationals, there is a sequence of rationals and irrationals converging to $a$. But, $lim_{x to a} f(x)$ exists if and only if for any two sequences $a_n, b_n to a$ we have $lim a_n = lim b_n$, and here the sequence of rationals and irrationals produce different values, namely $0$ and $1$.
answered Nov 21 '18 at 5:02
астон вілла олоф мэллбэрг
37.3k33376
Your proof is complete and fine. Only one minor correction : "hence $f$ would not get arbitrarily small" should be changed to "hence $f$ would not get arbitrarily close to the possible limit", in the first paragraph of the proof (or changed to "hence $f-L$ would not get arbitrarily small, where $L neq 0,1$ is a possible limit").
There is another proof that will avoid the breaking of $L$ into cases : indeed, given $a in mathbb R$, by density of rationals and irrationals, there is a sequence of rationals and irrationals converging to $a$. But, $lim_{x to a} f(x)$ exists if and only if for any two sequences $a_n, b_n to a$ we have $lim a_n = lim b_n$, and here the sequence of rationals and irrationals produce different values, namely $0$ and $1$.
answered Nov 21 '18 at 5:02
астон вілла олоф мэллбэрг
37.3k33376
answered Nov 21 '18 at 5:02
астон вілла олоф мэллбэрг
37.3k33376
answered Nov 21 '18 at 5:02
астон вілла олоф мэллбэрг
37.3k33376
answered Nov 21 '18 at 5:02
астон вілла олоф мэллбэрг
37.3k33376
37.3k33376
add a comment |
add a comment |
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Exactly what are you asking? What point is the one you are unsure about?
– Will M.
Nov 21 '18 at 4:13
I am unsure of whether my proof is correct. For example, is assuming that the only possible limits are 0 and 1 a correct assumption? Then do I correctly rule these out as limits?
– Wesley
Nov 21 '18 at 4:15
Ok, technically speaking, you need to consider a value, whatever it may be, $L in mathbf{R},$ and then rule it out as a possible limit. You can divide three cases: (1) $L neq 0, 1$ (2) $L = 0$ and (3) $L=1.$
– Will M.
Nov 21 '18 at 4:17
Is at least my reasoning for cases (2) and (3) correct?
– Wesley
Nov 21 '18 at 4:30