Proof for continuity of $h(x) = begin{cases} x, & xinmathbb{Q} \ -x, & xnotinmathbb{Q} end{cases}$
$begingroup$
I have the given function and I have to find out, where it is continuous and discontinuous...
My guess is, that the function is discontinuous on all $mathbb{R}setminus{0}$ and continuous in $0$. Could you look over my proof and give me a feedback?
Proof:
For $xinmathbb{Q}$ look at the sequence $(q_n)$, which converges to $x$. The sequence $(h(q_n))$ is then equal to the sequence $(q_n)$ and therefore converges to $x$, so we can conclude: $lim_{n->infty}h(q_n))=h(lim_{n->infty}q_n)=h(x)=x$
For $xnotinmathbb{Q}$ look at the sequence $(r_n)$, which converges to $x$. The sequence $(h(r_n))$ is then equal to the sequence $-(r_n)$ and therefore converges to $-x$, so we can conclude: $lim_{n->infty}h(r_n))=xneq-x=h(lim_{n->infty}-r_n)$
Therefore $h$ can't be continuous (I don't know the exact theorem name, but the subsequences of a function has to have the limit $h(x_0)$, if $h(x)$ is continuous in $x_0$
So the function is only continuous in $x=0$. Proof:
Let $epsilon>0$ and $delta =epsilon$, then we have:
$forall epsilon>0 exists delta > 0 forall xin mathbb{R} : |x-x_0|=|x-0|<delta =epsilon Rightarrow |h(x)-h(x_0)|=|h(x)-x_0|=|h(x)|=|x|<epsilon$
We can conclude, that the function is continuous in 0 and discontinuous for every $mathbb{R} setminus 0$
real-analysis calculus continuity
asked Dec 8 '18 at 18:59
BekayBekay
212
$endgroup$
|
show 6 more comments
$begingroup$
I have the given function and I have to find out, where it is continuous and discontinuous...
My guess is, that the function is discontinuous on all $mathbb{R}setminus{0}$ and continuous in $0$. Could you look over my proof and give me a feedback?
Proof:
For $xinmathbb{Q}$ look at the sequence $(q_n)$, which converges to $x$. The sequence $(h(q_n))$ is then equal to the sequence $(q_n)$ and therefore converges to $x$, so we can conclude: $lim_{n->infty}h(q_n))=h(lim_{n->infty}q_n)=h(x)=x$
For $xnotinmathbb{Q}$ look at the sequence $(r_n)$, which converges to $x$. The sequence $(h(r_n))$ is then equal to the sequence $-(r_n)$ and therefore converges to $-x$, so we can conclude: $lim_{n->infty}h(r_n))=xneq-x=h(lim_{n->infty}-r_n)$
Therefore $h$ can't be continuous (I don't know the exact theorem name, but the subsequences of a function has to have the limit $h(x_0)$, if $h(x)$ is continuous in $x_0$
So the function is only continuous in $x=0$. Proof:
Let $epsilon>0$ and $delta =epsilon$, then we have:
$forall epsilon>0 exists delta > 0 forall xin mathbb{R} : |x-x_0|=|x-0|<delta =epsilon Rightarrow |h(x)-h(x_0)|=|h(x)-x_0|=|h(x)|=|x|<epsilon$
We can conclude, that the function is continuous in 0 and discontinuous for every $mathbb{R} setminus 0$
real-analysis calculus continuity
asked Dec 8 '18 at 18:59
BekayBekay
212
$endgroup$
1
$begingroup$
Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:09
$begingroup$
By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
$endgroup$
– Bekay
Dec 8 '18 at 19:21
$begingroup$
I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:28
$begingroup$
Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
$endgroup$
– Bekay
Dec 8 '18 at 19:38
$begingroup$
It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:43
|
show 6 more comments
$begingroup$
I have the given function and I have to find out, where it is continuous and discontinuous...
My guess is, that the function is discontinuous on all $mathbb{R}setminus{0}$ and continuous in $0$. Could you look over my proof and give me a feedback?
Proof:
For $xinmathbb{Q}$ look at the sequence $(q_n)$, which converges to $x$. The sequence $(h(q_n))$ is then equal to the sequence $(q_n)$ and therefore converges to $x$, so we can conclude: $lim_{n->infty}h(q_n))=h(lim_{n->infty}q_n)=h(x)=x$
For $xnotinmathbb{Q}$ look at the sequence $(r_n)$, which converges to $x$. The sequence $(h(r_n))$ is then equal to the sequence $-(r_n)$ and therefore converges to $-x$, so we can conclude: $lim_{n->infty}h(r_n))=xneq-x=h(lim_{n->infty}-r_n)$
Therefore $h$ can't be continuous (I don't know the exact theorem name, but the subsequences of a function has to have the limit $h(x_0)$, if $h(x)$ is continuous in $x_0$
So the function is only continuous in $x=0$. Proof:
Let $epsilon>0$ and $delta =epsilon$, then we have:
$forall epsilon>0 exists delta > 0 forall xin mathbb{R} : |x-x_0|=|x-0|<delta =epsilon Rightarrow |h(x)-h(x_0)|=|h(x)-x_0|=|h(x)|=|x|<epsilon$
We can conclude, that the function is continuous in 0 and discontinuous for every $mathbb{R} setminus 0$
real-analysis calculus continuity
asked Dec 8 '18 at 18:59
BekayBekay
212
$endgroup$
I have the given function and I have to find out, where it is continuous and discontinuous...
My guess is, that the function is discontinuous on all $mathbb{R}setminus{0}$ and continuous in $0$. Could you look over my proof and give me a feedback?
Proof:
For $xinmathbb{Q}$ look at the sequence $(q_n)$, which converges to $x$. The sequence $(h(q_n))$ is then equal to the sequence $(q_n)$ and therefore converges to $x$, so we can conclude: $lim_{n->infty}h(q_n))=h(lim_{n->infty}q_n)=h(x)=x$
For $xnotinmathbb{Q}$ look at the sequence $(r_n)$, which converges to $x$. The sequence $(h(r_n))$ is then equal to the sequence $-(r_n)$ and therefore converges to $-x$, so we can conclude: $lim_{n->infty}h(r_n))=xneq-x=h(lim_{n->infty}-r_n)$
Therefore $h$ can't be continuous (I don't know the exact theorem name, but the subsequences of a function has to have the limit $h(x_0)$, if $h(x)$ is continuous in $x_0$
So the function is only continuous in $x=0$. Proof:
Let $epsilon>0$ and $delta =epsilon$, then we have:
$forall epsilon>0 exists delta > 0 forall xin mathbb{R} : |x-x_0|=|x-0|<delta =epsilon Rightarrow |h(x)-h(x_0)|=|h(x)-x_0|=|h(x)|=|x|<epsilon$
We can conclude, that the function is continuous in 0 and discontinuous for every $mathbb{R} setminus 0$
real-analysis calculus continuity
real-analysis calculus continuity
asked Dec 8 '18 at 18:59
BekayBekay
212
asked Dec 8 '18 at 18:59
BekayBekay
212
edited Dec 8 '18 at 19:37
Bekay
asked Dec 8 '18 at 18:59
BekayBekay
212
asked Dec 8 '18 at 18:59
BekayBekay
212
asked Dec 8 '18 at 18:59
BekayBekay
212
212
1
$begingroup$
Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:09
$begingroup$
By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
$endgroup$
– Bekay
Dec 8 '18 at 19:21
$begingroup$
I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:28
$begingroup$
Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
$endgroup$
– Bekay
Dec 8 '18 at 19:38
$begingroup$
It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:43
|
show 6 more comments
1
$begingroup$
Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:09
$begingroup$
By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
$endgroup$
– Bekay
Dec 8 '18 at 19:21
$begingroup$
I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:28
$begingroup$
Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
$endgroup$
– Bekay
Dec 8 '18 at 19:38
$begingroup$
It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:43
1
1
$begingroup$
Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:09
$begingroup$
Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:09
$begingroup$
By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
$endgroup$
– Bekay
Dec 8 '18 at 19:21
$begingroup$
By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
$endgroup$
– Bekay
Dec 8 '18 at 19:21
$begingroup$
I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:28
$begingroup$
I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:28
$begingroup$
Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
$endgroup$
– Bekay
Dec 8 '18 at 19:38
$begingroup$
Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
$endgroup$
– Bekay
Dec 8 '18 at 19:38
$begingroup$
It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:43
$begingroup$
It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:43
|
show 6 more comments
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$begingroup$
Your guess is correct. The proof for continuity in $0$ is also good, but you can omit the "$forall epsilon > 0 exists delta > 0$", because you already chose an arbitrary $epsilon$ and $delta=epsilon$. The proof of discontinuity in any $x in mathbb{R} setminus {0}$ is not quite correct. If you pick a sequence $x_n to x$, the $x_n$ could be both rational or irrational numbers. You should try to pick a specific sequence $x_n to x$ such that $h(x_n) to x$ is not true.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:09
$begingroup$
By specific you mean something like $x_n = frac{1}{n}$ (just an example, not a valid sequence :P)?
$endgroup$
– Bekay
Dec 8 '18 at 19:21
$begingroup$
I mean that, if you want to show discontinuity in $x$, it suffices to show that there exists a sequence $x_n to x$ such that $h(x_n) to h(x)$ doesn't hold. That doesn't mean that you have to be able to write the sequence as specifically as "$x_n = frac1n$." Btw, in the proof of continuity in $0$, it should be $h$ instead of $f$.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:28
$begingroup$
Ohh yeah, thanks for the $h$. But what kind of sequence would that be? Could you give an example?
$endgroup$
– Bekay
Dec 8 '18 at 19:38
$begingroup$
It could be anything, so I can't give an example. We just need to know that a series with these properties exists, we don't have to know what it is exactly.
$endgroup$
– Tki Deneb
Dec 8 '18 at 19:43