Continuity of a piecewise defined function
$begingroup$
The function f(x) is defined by:
f(x)=x, if x is rational
f(x)=2x, if x is irrational
Is f continuous?
Now my answer was the following: let a=√2 (any irrational number would do). Then f(a)=2√2.
But, as x gets arbitrarily close to a, x is rational, so lim(x->a) of f(x)=√2, which is not equal to f(a), so f is not continuous.
Are there any flaws in my proof? My main question is, can we safely assume that x is rational as we let it get arbitrarily close (but not equal) to a?
Thanks in advance, and excuse my failure to use MathJax, but I cannot for the love of me get the mathematical notations to appear.
continuity
asked Dec 17 '18 at 18:43
JBuckJBuck
566
$endgroup$
add a comment |
$begingroup$
The function f(x) is defined by:
f(x)=x, if x is rational
f(x)=2x, if x is irrational
Is f continuous?
Now my answer was the following: let a=√2 (any irrational number would do). Then f(a)=2√2.
But, as x gets arbitrarily close to a, x is rational, so lim(x->a) of f(x)=√2, which is not equal to f(a), so f is not continuous.
Are there any flaws in my proof? My main question is, can we safely assume that x is rational as we let it get arbitrarily close (but not equal) to a?
Thanks in advance, and excuse my failure to use MathJax, but I cannot for the love of me get the mathematical notations to appear.
continuity
asked Dec 17 '18 at 18:43
JBuckJBuck
566
$endgroup$
add a comment |
$begingroup$
The function f(x) is defined by:
f(x)=x, if x is rational
f(x)=2x, if x is irrational
Is f continuous?
Now my answer was the following: let a=√2 (any irrational number would do). Then f(a)=2√2.
But, as x gets arbitrarily close to a, x is rational, so lim(x->a) of f(x)=√2, which is not equal to f(a), so f is not continuous.
Are there any flaws in my proof? My main question is, can we safely assume that x is rational as we let it get arbitrarily close (but not equal) to a?
Thanks in advance, and excuse my failure to use MathJax, but I cannot for the love of me get the mathematical notations to appear.
continuity
asked Dec 17 '18 at 18:43
JBuckJBuck
566
$endgroup$
The function f(x) is defined by:
f(x)=x, if x is rational
f(x)=2x, if x is irrational
Is f continuous?
Now my answer was the following: let a=√2 (any irrational number would do). Then f(a)=2√2.
But, as x gets arbitrarily close to a, x is rational, so lim(x->a) of f(x)=√2, which is not equal to f(a), so f is not continuous.
Are there any flaws in my proof? My main question is, can we safely assume that x is rational as we let it get arbitrarily close (but not equal) to a?
Thanks in advance, and excuse my failure to use MathJax, but I cannot for the love of me get the mathematical notations to appear.
continuity
continuity
asked Dec 17 '18 at 18:43
JBuckJBuck
566
asked Dec 17 '18 at 18:43
JBuckJBuck
566
asked Dec 17 '18 at 18:43
JBuckJBuck
566
asked Dec 17 '18 at 18:43
JBuckJBuck
566
asked Dec 17 '18 at 18:43
JBuckJBuck
566
566
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your proof is not quite correct. In particular, the statement "as $x$ gets arbitrarily close to $a$, $x$ is rational" is not correct. In fact, no matter how close $x$ gets to $a$, there will always be rational and irrational numbers between $x$ and $a$.
For $f$ to be continuous at a point $x=a,$ we would need $lim_{xto a}f(x)=f(a)$. That means that for all $epsilon > 0$, there must be a $delta>0$ such that if $xin(a-delta, a+delta)$, $f(x)in (f(a)-epsilon, f(a)+epsilon)$. This clearly cannot happen in this case because any interval $(a-delta,a+delta)$ contains rational and irrational numbers, so the values of $f(x)$ cannot all be within $epsilon$ of $f(a)$ (whether $a$ is rational or irrational).
Let's see what this would look like for your choice of $a= sqrt{2}$ with $f(a)=2sqrt{2}$. Let $epsilon=sqrt{2}$. For any $delta>0$, the interval $(a-delta,a)$ contains a rational number, call it $q$. Then
$$f(a)-f(q)=2sqrt{2}-q>2sqrt{2}-sqrt{2}=sqrt{2}=epsilon$$
answered Dec 17 '18 at 19:07
pwerthpwerth
2,898416
$endgroup$
$begingroup$
Thank you! Unfortunately I haven't yet been taught this definition of continuity, therefore I get the general idea of your proof, but it will require extra effort and serious thinking on my part to grasp all the little details.
$endgroup$
– JBuck
Dec 17 '18 at 19:36
$begingroup$
What definition of continuity have you seen? I just used the standard $epsilon-delta$ definition here.
$endgroup$
– pwerth
Dec 17 '18 at 19:44
$begingroup$
The definition that was given to us said that if the limit of f as x approaches a exists and is equal to f(a), then f is continuous at a. Moreover, the definition of a limit we were given was "in words" and did not include any of the formal notation with ε, δ and such.
$endgroup$
– JBuck
Dec 17 '18 at 20:51
$begingroup$
As a side note, I now think I have the grasp of the proof, except for one thing: "...so the values of f(x) cannot all be within ϵ of f(a) (whether a is rational or irrational)" . By that, you mean that, for this particular function, no matter the ε, δ, there will always be a number that is outside the specified interval?
$endgroup$
– JBuck
Dec 17 '18 at 20:58
$begingroup$
What does it mean to say that the limit of $f$ as $x$ approaches $a$ exists and is equal to $f(a)$? If you write this out, this will be precisely the conditions I wrote! And yes, what I mean is that if $epsilon<sqrt{2}$, there is no way to ensure that all function values are within $epsilon$ of $f(a)=f(sqrt{2})$ because of the way the function is defined.
$endgroup$
– pwerth
Dec 18 '18 at 5:11
add a comment |
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$begingroup$
Your proof is not quite correct. In particular, the statement "as $x$ gets arbitrarily close to $a$, $x$ is rational" is not correct. In fact, no matter how close $x$ gets to $a$, there will always be rational and irrational numbers between $x$ and $a$.
For $f$ to be continuous at a point $x=a,$ we would need $lim_{xto a}f(x)=f(a)$. That means that for all $epsilon > 0$, there must be a $delta>0$ such that if $xin(a-delta, a+delta)$, $f(x)in (f(a)-epsilon, f(a)+epsilon)$. This clearly cannot happen in this case because any interval $(a-delta,a+delta)$ contains rational and irrational numbers, so the values of $f(x)$ cannot all be within $epsilon$ of $f(a)$ (whether $a$ is rational or irrational).
Let's see what this would look like for your choice of $a= sqrt{2}$ with $f(a)=2sqrt{2}$. Let $epsilon=sqrt{2}$. For any $delta>0$, the interval $(a-delta,a)$ contains a rational number, call it $q$. Then
$$f(a)-f(q)=2sqrt{2}-q>2sqrt{2}-sqrt{2}=sqrt{2}=epsilon$$
answered Dec 17 '18 at 19:07
pwerthpwerth
2,898416
$endgroup$
$begingroup$
Thank you! Unfortunately I haven't yet been taught this definition of continuity, therefore I get the general idea of your proof, but it will require extra effort and serious thinking on my part to grasp all the little details.
$endgroup$
– JBuck
Dec 17 '18 at 19:36
$begingroup$
What definition of continuity have you seen? I just used the standard $epsilon-delta$ definition here.
$endgroup$
– pwerth
Dec 17 '18 at 19:44
$begingroup$
The definition that was given to us said that if the limit of f as x approaches a exists and is equal to f(a), then f is continuous at a. Moreover, the definition of a limit we were given was "in words" and did not include any of the formal notation with ε, δ and such.
$endgroup$
– JBuck
Dec 17 '18 at 20:51
$begingroup$
As a side note, I now think I have the grasp of the proof, except for one thing: "...so the values of f(x) cannot all be within ϵ of f(a) (whether a is rational or irrational)" . By that, you mean that, for this particular function, no matter the ε, δ, there will always be a number that is outside the specified interval?
$endgroup$
– JBuck
Dec 17 '18 at 20:58
$begingroup$
What does it mean to say that the limit of $f$ as $x$ approaches $a$ exists and is equal to $f(a)$? If you write this out, this will be precisely the conditions I wrote! And yes, what I mean is that if $epsilon<sqrt{2}$, there is no way to ensure that all function values are within $epsilon$ of $f(a)=f(sqrt{2})$ because of the way the function is defined.
$endgroup$
– pwerth
Dec 18 '18 at 5:11
add a comment |
$begingroup$
Your proof is not quite correct. In particular, the statement "as $x$ gets arbitrarily close to $a$, $x$ is rational" is not correct. In fact, no matter how close $x$ gets to $a$, there will always be rational and irrational numbers between $x$ and $a$.
For $f$ to be continuous at a point $x=a,$ we would need $lim_{xto a}f(x)=f(a)$. That means that for all $epsilon > 0$, there must be a $delta>0$ such that if $xin(a-delta, a+delta)$, $f(x)in (f(a)-epsilon, f(a)+epsilon)$. This clearly cannot happen in this case because any interval $(a-delta,a+delta)$ contains rational and irrational numbers, so the values of $f(x)$ cannot all be within $epsilon$ of $f(a)$ (whether $a$ is rational or irrational).
Let's see what this would look like for your choice of $a= sqrt{2}$ with $f(a)=2sqrt{2}$. Let $epsilon=sqrt{2}$. For any $delta>0$, the interval $(a-delta,a)$ contains a rational number, call it $q$. Then
$$f(a)-f(q)=2sqrt{2}-q>2sqrt{2}-sqrt{2}=sqrt{2}=epsilon$$
answered Dec 17 '18 at 19:07
pwerthpwerth
2,898416
$endgroup$
$begingroup$
Thank you! Unfortunately I haven't yet been taught this definition of continuity, therefore I get the general idea of your proof, but it will require extra effort and serious thinking on my part to grasp all the little details.
$endgroup$
– JBuck
Dec 17 '18 at 19:36
$begingroup$
What definition of continuity have you seen? I just used the standard $epsilon-delta$ definition here.
$endgroup$
– pwerth
Dec 17 '18 at 19:44
$begingroup$
The definition that was given to us said that if the limit of f as x approaches a exists and is equal to f(a), then f is continuous at a. Moreover, the definition of a limit we were given was "in words" and did not include any of the formal notation with ε, δ and such.
$endgroup$
– JBuck
Dec 17 '18 at 20:51
$begingroup$
As a side note, I now think I have the grasp of the proof, except for one thing: "...so the values of f(x) cannot all be within ϵ of f(a) (whether a is rational or irrational)" . By that, you mean that, for this particular function, no matter the ε, δ, there will always be a number that is outside the specified interval?
$endgroup$
– JBuck
Dec 17 '18 at 20:58
$begingroup$
What does it mean to say that the limit of $f$ as $x$ approaches $a$ exists and is equal to $f(a)$? If you write this out, this will be precisely the conditions I wrote! And yes, what I mean is that if $epsilon<sqrt{2}$, there is no way to ensure that all function values are within $epsilon$ of $f(a)=f(sqrt{2})$ because of the way the function is defined.
$endgroup$
– pwerth
Dec 18 '18 at 5:11
add a comment |
$begingroup$
Your proof is not quite correct. In particular, the statement "as $x$ gets arbitrarily close to $a$, $x$ is rational" is not correct. In fact, no matter how close $x$ gets to $a$, there will always be rational and irrational numbers between $x$ and $a$.
For $f$ to be continuous at a point $x=a,$ we would need $lim_{xto a}f(x)=f(a)$. That means that for all $epsilon > 0$, there must be a $delta>0$ such that if $xin(a-delta, a+delta)$, $f(x)in (f(a)-epsilon, f(a)+epsilon)$. This clearly cannot happen in this case because any interval $(a-delta,a+delta)$ contains rational and irrational numbers, so the values of $f(x)$ cannot all be within $epsilon$ of $f(a)$ (whether $a$ is rational or irrational).
Let's see what this would look like for your choice of $a= sqrt{2}$ with $f(a)=2sqrt{2}$. Let $epsilon=sqrt{2}$. For any $delta>0$, the interval $(a-delta,a)$ contains a rational number, call it $q$. Then
$$f(a)-f(q)=2sqrt{2}-q>2sqrt{2}-sqrt{2}=sqrt{2}=epsilon$$
answered Dec 17 '18 at 19:07
pwerthpwerth
2,898416
$endgroup$
Your proof is not quite correct. In particular, the statement "as $x$ gets arbitrarily close to $a$, $x$ is rational" is not correct. In fact, no matter how close $x$ gets to $a$, there will always be rational and irrational numbers between $x$ and $a$.
For $f$ to be continuous at a point $x=a,$ we would need $lim_{xto a}f(x)=f(a)$. That means that for all $epsilon > 0$, there must be a $delta>0$ such that if $xin(a-delta, a+delta)$, $f(x)in (f(a)-epsilon, f(a)+epsilon)$. This clearly cannot happen in this case because any interval $(a-delta,a+delta)$ contains rational and irrational numbers, so the values of $f(x)$ cannot all be within $epsilon$ of $f(a)$ (whether $a$ is rational or irrational).
Let's see what this would look like for your choice of $a= sqrt{2}$ with $f(a)=2sqrt{2}$. Let $epsilon=sqrt{2}$. For any $delta>0$, the interval $(a-delta,a)$ contains a rational number, call it $q$. Then
$$f(a)-f(q)=2sqrt{2}-q>2sqrt{2}-sqrt{2}=sqrt{2}=epsilon$$
answered Dec 17 '18 at 19:07
pwerthpwerth
2,898416
answered Dec 17 '18 at 19:07
pwerthpwerth
2,898416
answered Dec 17 '18 at 19:07
pwerthpwerth
2,898416
answered Dec 17 '18 at 19:07
pwerthpwerth
2,898416
2,898416
$begingroup$
Thank you! Unfortunately I haven't yet been taught this definition of continuity, therefore I get the general idea of your proof, but it will require extra effort and serious thinking on my part to grasp all the little details.
$endgroup$
– JBuck
Dec 17 '18 at 19:36
$begingroup$
What definition of continuity have you seen? I just used the standard $epsilon-delta$ definition here.
$endgroup$
– pwerth
Dec 17 '18 at 19:44
$begingroup$
The definition that was given to us said that if the limit of f as x approaches a exists and is equal to f(a), then f is continuous at a. Moreover, the definition of a limit we were given was "in words" and did not include any of the formal notation with ε, δ and such.
$endgroup$
– JBuck
Dec 17 '18 at 20:51
$begingroup$
As a side note, I now think I have the grasp of the proof, except for one thing: "...so the values of f(x) cannot all be within ϵ of f(a) (whether a is rational or irrational)" . By that, you mean that, for this particular function, no matter the ε, δ, there will always be a number that is outside the specified interval?
$endgroup$
– JBuck
Dec 17 '18 at 20:58
$begingroup$
What does it mean to say that the limit of $f$ as $x$ approaches $a$ exists and is equal to $f(a)$? If you write this out, this will be precisely the conditions I wrote! And yes, what I mean is that if $epsilon<sqrt{2}$, there is no way to ensure that all function values are within $epsilon$ of $f(a)=f(sqrt{2})$ because of the way the function is defined.
$endgroup$
– pwerth
Dec 18 '18 at 5:11
add a comment |
$begingroup$
Thank you! Unfortunately I haven't yet been taught this definition of continuity, therefore I get the general idea of your proof, but it will require extra effort and serious thinking on my part to grasp all the little details.
$endgroup$
– JBuck
Dec 17 '18 at 19:36
$begingroup$
What definition of continuity have you seen? I just used the standard $epsilon-delta$ definition here.
$endgroup$
– pwerth
Dec 17 '18 at 19:44
$begingroup$
The definition that was given to us said that if the limit of f as x approaches a exists and is equal to f(a), then f is continuous at a. Moreover, the definition of a limit we were given was "in words" and did not include any of the formal notation with ε, δ and such.
$endgroup$
– JBuck
Dec 17 '18 at 20:51
$begingroup$
As a side note, I now think I have the grasp of the proof, except for one thing: "...so the values of f(x) cannot all be within ϵ of f(a) (whether a is rational or irrational)" . By that, you mean that, for this particular function, no matter the ε, δ, there will always be a number that is outside the specified interval?
$endgroup$
– JBuck
Dec 17 '18 at 20:58
$begingroup$
What does it mean to say that the limit of $f$ as $x$ approaches $a$ exists and is equal to $f(a)$? If you write this out, this will be precisely the conditions I wrote! And yes, what I mean is that if $epsilon<sqrt{2}$, there is no way to ensure that all function values are within $epsilon$ of $f(a)=f(sqrt{2})$ because of the way the function is defined.
$endgroup$
– pwerth
Dec 18 '18 at 5:11
$begingroup$
Thank you! Unfortunately I haven't yet been taught this definition of continuity, therefore I get the general idea of your proof, but it will require extra effort and serious thinking on my part to grasp all the little details.
$endgroup$
– JBuck
Dec 17 '18 at 19:36
$begingroup$
Thank you! Unfortunately I haven't yet been taught this definition of continuity, therefore I get the general idea of your proof, but it will require extra effort and serious thinking on my part to grasp all the little details.
$endgroup$
– JBuck
Dec 17 '18 at 19:36
$begingroup$
What definition of continuity have you seen? I just used the standard $epsilon-delta$ definition here.
$endgroup$
– pwerth
Dec 17 '18 at 19:44
$begingroup$
What definition of continuity have you seen? I just used the standard $epsilon-delta$ definition here.
$endgroup$
– pwerth
Dec 17 '18 at 19:44
$begingroup$
The definition that was given to us said that if the limit of f as x approaches a exists and is equal to f(a), then f is continuous at a. Moreover, the definition of a limit we were given was "in words" and did not include any of the formal notation with ε, δ and such.
$endgroup$
– JBuck
Dec 17 '18 at 20:51
$begingroup$
The definition that was given to us said that if the limit of f as x approaches a exists and is equal to f(a), then f is continuous at a. Moreover, the definition of a limit we were given was "in words" and did not include any of the formal notation with ε, δ and such.
$endgroup$
– JBuck
Dec 17 '18 at 20:51
$begingroup$
As a side note, I now think I have the grasp of the proof, except for one thing: "...so the values of f(x) cannot all be within ϵ of f(a) (whether a is rational or irrational)" . By that, you mean that, for this particular function, no matter the ε, δ, there will always be a number that is outside the specified interval?
$endgroup$
– JBuck
Dec 17 '18 at 20:58
$begingroup$
As a side note, I now think I have the grasp of the proof, except for one thing: "...so the values of f(x) cannot all be within ϵ of f(a) (whether a is rational or irrational)" . By that, you mean that, for this particular function, no matter the ε, δ, there will always be a number that is outside the specified interval?
$endgroup$
– JBuck
Dec 17 '18 at 20:58
$begingroup$
What does it mean to say that the limit of $f$ as $x$ approaches $a$ exists and is equal to $f(a)$? If you write this out, this will be precisely the conditions I wrote! And yes, what I mean is that if $epsilon<sqrt{2}$, there is no way to ensure that all function values are within $epsilon$ of $f(a)=f(sqrt{2})$ because of the way the function is defined.
$endgroup$
– pwerth
Dec 18 '18 at 5:11
$begingroup$
What does it mean to say that the limit of $f$ as $x$ approaches $a$ exists and is equal to $f(a)$? If you write this out, this will be precisely the conditions I wrote! And yes, what I mean is that if $epsilon<sqrt{2}$, there is no way to ensure that all function values are within $epsilon$ of $f(a)=f(sqrt{2})$ because of the way the function is defined.
$endgroup$
– pwerth
Dec 18 '18 at 5:11
add a comment |
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