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Prove or disprove: If $(x_n)$ is a Cauchy sequence in $X$ and $f$ is a continuous function on $X$, then $(f(x_n))$ is a Cauchy sequence.
Clash Royale CLAN TAG#URR8PPP
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Note that $X$ is a metric space and $f:(X,rho)rightarrow (X,sigma)$.
Disproving: Suppose $(x_n)$ is a Cauchy sequence, $f$ is continuous and $(f(x_n))$ is not a Cauchy sequence, then $existsepsilon'>0$ such that $forall MinBbbN$, there exists $i,jgeq M$ where $sigma(f(x_i),f(x_j))>epsilon'$. Also, knowing that $f$ is continuous, then we know that $existsdelta'>0$ such that if $rho(x_i,y)<delta'$, such that $yin X$, then $sigma(f(x_i),f(y))<epsilon'$. However, $y$ is not necessarily a term of the Cauchy sequence $(x_n)$. Hence, it does not necessarily mean that $(f(x_n))$ is also a Cauchy sequence.
Now, I just noticed after typing that this may get tagged as duplicate since the same question has been asked here: If $X = x_n:n in mathbb N$ is a cauchy sequence in a metric space $S$ and $f : S rightarrow T$ is continuous , is $f(x_n)$ a cauchy sequence?, but would my argument be valid in general? Thank you!
real-analysis metric-spaces continuity cauchy-sequences
asked Jul 18 at 13:56
TheLast Cipher
538414
 |Â
show 5 more comments
up vote
2
down vote
favorite
Note that $X$ is a metric space and $f:(X,rho)rightarrow (X,sigma)$.
Disproving: Suppose $(x_n)$ is a Cauchy sequence, $f$ is continuous and $(f(x_n))$ is not a Cauchy sequence, then $existsepsilon'>0$ such that $forall MinBbbN$, there exists $i,jgeq M$ where $sigma(f(x_i),f(x_j))>epsilon'$. Also, knowing that $f$ is continuous, then we know that $existsdelta'>0$ such that if $rho(x_i,y)<delta'$, such that $yin X$, then $sigma(f(x_i),f(y))<epsilon'$. However, $y$ is not necessarily a term of the Cauchy sequence $(x_n)$. Hence, it does not necessarily mean that $(f(x_n))$ is also a Cauchy sequence.
Now, I just noticed after typing that this may get tagged as duplicate since the same question has been asked here: If $X = x_n:n in mathbb N$ is a cauchy sequence in a metric space $S$ and $f : S rightarrow T$ is continuous , is $f(x_n)$ a cauchy sequence?, but would my argument be valid in general? Thank you!
real-analysis metric-spaces continuity cauchy-sequences
asked Jul 18 at 13:56
TheLast Cipher
538414
1
How do you choose your $y$ for it to work? And it should rather be $sigma (f(x_i),f(y))<epsilon '$ right?
– Suzet
Jul 18 at 13:58
What do you mean? I am not trying to make it work because I'm arguing it would not always work.
– TheLast Cipher
Jul 18 at 14:00
oh right. typo! thanks
– TheLast Cipher
Jul 18 at 14:01
I've corrected it now.
– TheLast Cipher
Jul 18 at 14:01
3
Oh, actually the statement is untrue, I didn't get it. Well, the only way to show that it is false is to give a counter example for it. Trying to explain why a proof of it wouldn't work is good, but you do not get a definitive conclusion before you give a counter example.
– Suzet
Jul 18 at 14:03
 |Â
show 5 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Note that $X$ is a metric space and $f:(X,rho)rightarrow (X,sigma)$.
Disproving: Suppose $(x_n)$ is a Cauchy sequence, $f$ is continuous and $(f(x_n))$ is not a Cauchy sequence, then $existsepsilon'>0$ such that $forall MinBbbN$, there exists $i,jgeq M$ where $sigma(f(x_i),f(x_j))>epsilon'$. Also, knowing that $f$ is continuous, then we know that $existsdelta'>0$ such that if $rho(x_i,y)<delta'$, such that $yin X$, then $sigma(f(x_i),f(y))<epsilon'$. However, $y$ is not necessarily a term of the Cauchy sequence $(x_n)$. Hence, it does not necessarily mean that $(f(x_n))$ is also a Cauchy sequence.
Now, I just noticed after typing that this may get tagged as duplicate since the same question has been asked here: If $X = x_n:n in mathbb N$ is a cauchy sequence in a metric space $S$ and $f : S rightarrow T$ is continuous , is $f(x_n)$ a cauchy sequence?, but would my argument be valid in general? Thank you!
real-analysis metric-spaces continuity cauchy-sequences
asked Jul 18 at 13:56
TheLast Cipher
538414
Note that $X$ is a metric space and $f:(X,rho)rightarrow (X,sigma)$.
Disproving: Suppose $(x_n)$ is a Cauchy sequence, $f$ is continuous and $(f(x_n))$ is not a Cauchy sequence, then $existsepsilon'>0$ such that $forall MinBbbN$, there exists $i,jgeq M$ where $sigma(f(x_i),f(x_j))>epsilon'$. Also, knowing that $f$ is continuous, then we know that $existsdelta'>0$ such that if $rho(x_i,y)<delta'$, such that $yin X$, then $sigma(f(x_i),f(y))<epsilon'$. However, $y$ is not necessarily a term of the Cauchy sequence $(x_n)$. Hence, it does not necessarily mean that $(f(x_n))$ is also a Cauchy sequence.
Now, I just noticed after typing that this may get tagged as duplicate since the same question has been asked here: If $X = x_n:n in mathbb N$ is a cauchy sequence in a metric space $S$ and $f : S rightarrow T$ is continuous , is $f(x_n)$ a cauchy sequence?, but would my argument be valid in general? Thank you!
real-analysis metric-spaces continuity cauchy-sequences
asked Jul 18 at 13:56
TheLast Cipher
538414
edited Jul 18 at 14:20
asked Jul 18 at 13:56
TheLast Cipher
538414
asked Jul 18 at 13:56
TheLast Cipher
538414
asked Jul 18 at 13:56
TheLast Cipher
538414
538414
1
How do you choose your $y$ for it to work? And it should rather be $sigma (f(x_i),f(y))<epsilon '$ right?
– Suzet
Jul 18 at 13:58
What do you mean? I am not trying to make it work because I'm arguing it would not always work.
– TheLast Cipher
Jul 18 at 14:00
oh right. typo! thanks
– TheLast Cipher
Jul 18 at 14:01
I've corrected it now.
– TheLast Cipher
Jul 18 at 14:01
3
Oh, actually the statement is untrue, I didn't get it. Well, the only way to show that it is false is to give a counter example for it. Trying to explain why a proof of it wouldn't work is good, but you do not get a definitive conclusion before you give a counter example.
– Suzet
Jul 18 at 14:03
 |Â
show 5 more comments
1
How do you choose your $y$ for it to work? And it should rather be $sigma (f(x_i),f(y))<epsilon '$ right?
– Suzet
Jul 18 at 13:58
What do you mean? I am not trying to make it work because I'm arguing it would not always work.
– TheLast Cipher
Jul 18 at 14:00
oh right. typo! thanks
– TheLast Cipher
Jul 18 at 14:01
I've corrected it now.
– TheLast Cipher
Jul 18 at 14:01
3
Oh, actually the statement is untrue, I didn't get it. Well, the only way to show that it is false is to give a counter example for it. Trying to explain why a proof of it wouldn't work is good, but you do not get a definitive conclusion before you give a counter example.
– Suzet
Jul 18 at 14:03
1
1
How do you choose your $y$ for it to work? And it should rather be $sigma (f(x_i),f(y))<epsilon '$ right?
– Suzet
Jul 18 at 13:58
How do you choose your $y$ for it to work? And it should rather be $sigma (f(x_i),f(y))<epsilon '$ right?
– Suzet
Jul 18 at 13:58
What do you mean? I am not trying to make it work because I'm arguing it would not always work.
– TheLast Cipher
Jul 18 at 14:00
What do you mean? I am not trying to make it work because I'm arguing it would not always work.
– TheLast Cipher
Jul 18 at 14:00
oh right. typo! thanks
– TheLast Cipher
Jul 18 at 14:01
oh right. typo! thanks
– TheLast Cipher
Jul 18 at 14:01
I've corrected it now.
– TheLast Cipher
Jul 18 at 14:01
I've corrected it now.
– TheLast Cipher
Jul 18 at 14:01
3
3
Oh, actually the statement is untrue, I didn't get it. Well, the only way to show that it is false is to give a counter example for it. Trying to explain why a proof of it wouldn't work is good, but you do not get a definitive conclusion before you give a counter example.
– Suzet
Jul 18 at 14:03
Oh, actually the statement is untrue, I didn't get it. Well, the only way to show that it is false is to give a counter example for it. Trying to explain why a proof of it wouldn't work is good, but you do not get a definitive conclusion before you give a counter example.
– Suzet
Jul 18 at 14:03
 |Â
show 5 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Another counterexample that goes into a slightly different direction.
Choose $X=mathbb Q$,
$$f(x)= left{
beginmatrix
0 & text, if x < sqrt2\
1 & text, if x > sqrt2. \
endmatrix
right.
$$
This is a continuous function $mathbb Q to mathbb Q$.
If you choose a sequence $x_n$ of rationals that tends to $sqrt2$ from both sides (infinitely many terms both above and below $sqrt2$), then $x_n$ is Cauchy, but $f(x_n)$ is not, as it will contain infinitely many 0's and 1's.
The main idea is to realize that your proposition becomes true when $X$ is a complete metric space, so both my example and that of Salahamam Fatima use a non-complete space.
answered Jul 19 at 6:49
Ingix
2,205125
Interesting. I've proven the result when $X$ is a complete metric space, but somehow I still lacked intuition and struggle with finding counterexamples. Thank you for your input. This is informative.
– TheLast Cipher
Jul 19 at 7:25
That's why I posted my solution in addition to the already existing ones. Intuition about how things can 'break' from what one would expect from the 'normal' case comes from having seen lots of examples.
– Ingix
Jul 19 at 7:49
add a comment |Â
up vote
4
down vote
Take $$X=(0,+infty)$$
$$x_n=frac 1 n$$
$$f:xmapsto frac1x$$
$$(x_n) text is Cauchy$$
$$f text is continuous$$
$$f(x_n)=n text is not Cauchy$$
Your statement will be true if $f$ is UNIFORMLY CONTINUOUS.
answered Jul 18 at 14:10
Salahamam_ Fatima
33.6k21229
1
This does not provide any comment on the proof attempt, (like if it is correct or not).
– supinf
Jul 18 at 16:00
1
If you read his question, then you will notice he tried to disprove it too...
– supinf
Jul 18 at 16:04
@supinf The OP had to write $delta'_i$ instead of $delta'$ since it depends on $x_i$.
– Salahamam_ Fatima
Jul 18 at 16:09
@Salahamam_Fatima: I never wrote $delta_i'$. What do you mean it depends on $x_i$?
– TheLast Cipher
Jul 19 at 7:20
@TheLastCipher Yes.
– Salahamam_ Fatima
Jul 19 at 8:26
add a comment |Â
up vote
2
down vote
No, it is not correct. You start talking about $y$ without telling what $y$ is. But, above all, your approach only tells us why you think that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. The best approach, in this case, consists in providing a counterexample.
answered Jul 18 at 14:09
José Carlos Santos
114k1698177
I don't understand why arguing that something is not necessarily true is insufficient proof that it is not always true :? I have a hunch that this follow up falls under logic and philosophy? Does it?
– TheLast Cipher
Jul 18 at 14:34
What you said in your post was that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. That doesn't prove that there are cases in which it isn't a Cauchy sequence, which is what you were supposed to prove.
– José Carlos Santos
Jul 18 at 14:36
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Another counterexample that goes into a slightly different direction.
Choose $X=mathbb Q$,
$$f(x)= left{
beginmatrix
0 & text, if x < sqrt2\
1 & text, if x > sqrt2. \
endmatrix
right.
$$
This is a continuous function $mathbb Q to mathbb Q$.
If you choose a sequence $x_n$ of rationals that tends to $sqrt2$ from both sides (infinitely many terms both above and below $sqrt2$), then $x_n$ is Cauchy, but $f(x_n)$ is not, as it will contain infinitely many 0's and 1's.
The main idea is to realize that your proposition becomes true when $X$ is a complete metric space, so both my example and that of Salahamam Fatima use a non-complete space.
answered Jul 19 at 6:49
Ingix
2,205125
Interesting. I've proven the result when $X$ is a complete metric space, but somehow I still lacked intuition and struggle with finding counterexamples. Thank you for your input. This is informative.
– TheLast Cipher
Jul 19 at 7:25
That's why I posted my solution in addition to the already existing ones. Intuition about how things can 'break' from what one would expect from the 'normal' case comes from having seen lots of examples.
– Ingix
Jul 19 at 7:49
add a comment |Â
up vote
2
down vote
accepted
Another counterexample that goes into a slightly different direction.
Choose $X=mathbb Q$,
$$f(x)= left{
beginmatrix
0 & text, if x < sqrt2\
1 & text, if x > sqrt2. \
endmatrix
right.
$$
This is a continuous function $mathbb Q to mathbb Q$.
If you choose a sequence $x_n$ of rationals that tends to $sqrt2$ from both sides (infinitely many terms both above and below $sqrt2$), then $x_n$ is Cauchy, but $f(x_n)$ is not, as it will contain infinitely many 0's and 1's.
The main idea is to realize that your proposition becomes true when $X$ is a complete metric space, so both my example and that of Salahamam Fatima use a non-complete space.
answered Jul 19 at 6:49
Ingix
2,205125
Interesting. I've proven the result when $X$ is a complete metric space, but somehow I still lacked intuition and struggle with finding counterexamples. Thank you for your input. This is informative.
– TheLast Cipher
Jul 19 at 7:25
That's why I posted my solution in addition to the already existing ones. Intuition about how things can 'break' from what one would expect from the 'normal' case comes from having seen lots of examples.
– Ingix
Jul 19 at 7:49
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Another counterexample that goes into a slightly different direction.
Choose $X=mathbb Q$,
$$f(x)= left{
beginmatrix
0 & text, if x < sqrt2\
1 & text, if x > sqrt2. \
endmatrix
right.
$$
This is a continuous function $mathbb Q to mathbb Q$.
If you choose a sequence $x_n$ of rationals that tends to $sqrt2$ from both sides (infinitely many terms both above and below $sqrt2$), then $x_n$ is Cauchy, but $f(x_n)$ is not, as it will contain infinitely many 0's and 1's.
The main idea is to realize that your proposition becomes true when $X$ is a complete metric space, so both my example and that of Salahamam Fatima use a non-complete space.
answered Jul 19 at 6:49
Ingix
2,205125
Another counterexample that goes into a slightly different direction.
Choose $X=mathbb Q$,
$$f(x)= left{
beginmatrix
0 & text, if x < sqrt2\
1 & text, if x > sqrt2. \
endmatrix
right.
$$
This is a continuous function $mathbb Q to mathbb Q$.
If you choose a sequence $x_n$ of rationals that tends to $sqrt2$ from both sides (infinitely many terms both above and below $sqrt2$), then $x_n$ is Cauchy, but $f(x_n)$ is not, as it will contain infinitely many 0's and 1's.
The main idea is to realize that your proposition becomes true when $X$ is a complete metric space, so both my example and that of Salahamam Fatima use a non-complete space.
answered Jul 19 at 6:49
Ingix
2,205125
edited Jul 19 at 7:00
answered Jul 19 at 6:49
Ingix
2,205125
answered Jul 19 at 6:49
Ingix
2,205125
answered Jul 19 at 6:49
Ingix
2,205125
2,205125
Interesting. I've proven the result when $X$ is a complete metric space, but somehow I still lacked intuition and struggle with finding counterexamples. Thank you for your input. This is informative.
– TheLast Cipher
Jul 19 at 7:25
That's why I posted my solution in addition to the already existing ones. Intuition about how things can 'break' from what one would expect from the 'normal' case comes from having seen lots of examples.
– Ingix
Jul 19 at 7:49
add a comment |Â
Interesting. I've proven the result when $X$ is a complete metric space, but somehow I still lacked intuition and struggle with finding counterexamples. Thank you for your input. This is informative.
– TheLast Cipher
Jul 19 at 7:25
That's why I posted my solution in addition to the already existing ones. Intuition about how things can 'break' from what one would expect from the 'normal' case comes from having seen lots of examples.
– Ingix
Jul 19 at 7:49
Interesting. I've proven the result when $X$ is a complete metric space, but somehow I still lacked intuition and struggle with finding counterexamples. Thank you for your input. This is informative.
– TheLast Cipher
Jul 19 at 7:25
Interesting. I've proven the result when $X$ is a complete metric space, but somehow I still lacked intuition and struggle with finding counterexamples. Thank you for your input. This is informative.
– TheLast Cipher
Jul 19 at 7:25
That's why I posted my solution in addition to the already existing ones. Intuition about how things can 'break' from what one would expect from the 'normal' case comes from having seen lots of examples.
– Ingix
Jul 19 at 7:49
That's why I posted my solution in addition to the already existing ones. Intuition about how things can 'break' from what one would expect from the 'normal' case comes from having seen lots of examples.
– Ingix
Jul 19 at 7:49
add a comment |Â
up vote
4
down vote
Take $$X=(0,+infty)$$
$$x_n=frac 1 n$$
$$f:xmapsto frac1x$$
$$(x_n) text is Cauchy$$
$$f text is continuous$$
$$f(x_n)=n text is not Cauchy$$
Your statement will be true if $f$ is UNIFORMLY CONTINUOUS.
answered Jul 18 at 14:10
Salahamam_ Fatima
33.6k21229
1
This does not provide any comment on the proof attempt, (like if it is correct or not).
– supinf
Jul 18 at 16:00
1
If you read his question, then you will notice he tried to disprove it too...
– supinf
Jul 18 at 16:04
@supinf The OP had to write $delta'_i$ instead of $delta'$ since it depends on $x_i$.
– Salahamam_ Fatima
Jul 18 at 16:09
@Salahamam_Fatima: I never wrote $delta_i'$. What do you mean it depends on $x_i$?
– TheLast Cipher
Jul 19 at 7:20
@TheLastCipher Yes.
– Salahamam_ Fatima
Jul 19 at 8:26
add a comment |Â
up vote
4
down vote
Take $$X=(0,+infty)$$
$$x_n=frac 1 n$$
$$f:xmapsto frac1x$$
$$(x_n) text is Cauchy$$
$$f text is continuous$$
$$f(x_n)=n text is not Cauchy$$
Your statement will be true if $f$ is UNIFORMLY CONTINUOUS.
answered Jul 18 at 14:10
Salahamam_ Fatima
33.6k21229
1
This does not provide any comment on the proof attempt, (like if it is correct or not).
– supinf
Jul 18 at 16:00
1
If you read his question, then you will notice he tried to disprove it too...
– supinf
Jul 18 at 16:04
@supinf The OP had to write $delta'_i$ instead of $delta'$ since it depends on $x_i$.
– Salahamam_ Fatima
Jul 18 at 16:09
@Salahamam_Fatima: I never wrote $delta_i'$. What do you mean it depends on $x_i$?
– TheLast Cipher
Jul 19 at 7:20
@TheLastCipher Yes.
– Salahamam_ Fatima
Jul 19 at 8:26
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Take $$X=(0,+infty)$$
$$x_n=frac 1 n$$
$$f:xmapsto frac1x$$
$$(x_n) text is Cauchy$$
$$f text is continuous$$
$$f(x_n)=n text is not Cauchy$$
Your statement will be true if $f$ is UNIFORMLY CONTINUOUS.
answered Jul 18 at 14:10
Salahamam_ Fatima
33.6k21229
Take $$X=(0,+infty)$$
$$x_n=frac 1 n$$
$$f:xmapsto frac1x$$
$$(x_n) text is Cauchy$$
$$f text is continuous$$
$$f(x_n)=n text is not Cauchy$$
Your statement will be true if $f$ is UNIFORMLY CONTINUOUS.
answered Jul 18 at 14:10
Salahamam_ Fatima
33.6k21229
answered Jul 18 at 14:10
Salahamam_ Fatima
33.6k21229
answered Jul 18 at 14:10
Salahamam_ Fatima
33.6k21229
answered Jul 18 at 14:10
Salahamam_ Fatima
33.6k21229
33.6k21229
1
This does not provide any comment on the proof attempt, (like if it is correct or not).
– supinf
Jul 18 at 16:00
1
If you read his question, then you will notice he tried to disprove it too...
– supinf
Jul 18 at 16:04
@supinf The OP had to write $delta'_i$ instead of $delta'$ since it depends on $x_i$.
– Salahamam_ Fatima
Jul 18 at 16:09
@Salahamam_Fatima: I never wrote $delta_i'$. What do you mean it depends on $x_i$?
– TheLast Cipher
Jul 19 at 7:20
@TheLastCipher Yes.
– Salahamam_ Fatima
Jul 19 at 8:26
add a comment |Â
1
This does not provide any comment on the proof attempt, (like if it is correct or not).
– supinf
Jul 18 at 16:00
1
If you read his question, then you will notice he tried to disprove it too...
– supinf
Jul 18 at 16:04
@supinf The OP had to write $delta'_i$ instead of $delta'$ since it depends on $x_i$.
– Salahamam_ Fatima
Jul 18 at 16:09
@Salahamam_Fatima: I never wrote $delta_i'$. What do you mean it depends on $x_i$?
– TheLast Cipher
Jul 19 at 7:20
@TheLastCipher Yes.
– Salahamam_ Fatima
Jul 19 at 8:26
1
1
This does not provide any comment on the proof attempt, (like if it is correct or not).
– supinf
Jul 18 at 16:00
This does not provide any comment on the proof attempt, (like if it is correct or not).
– supinf
Jul 18 at 16:00
1
1
If you read his question, then you will notice he tried to disprove it too...
– supinf
Jul 18 at 16:04
If you read his question, then you will notice he tried to disprove it too...
– supinf
Jul 18 at 16:04
@supinf The OP had to write $delta'_i$ instead of $delta'$ since it depends on $x_i$.
– Salahamam_ Fatima
Jul 18 at 16:09
@supinf The OP had to write $delta'_i$ instead of $delta'$ since it depends on $x_i$.
– Salahamam_ Fatima
Jul 18 at 16:09
@Salahamam_Fatima: I never wrote $delta_i'$. What do you mean it depends on $x_i$?
– TheLast Cipher
Jul 19 at 7:20
@Salahamam_Fatima: I never wrote $delta_i'$. What do you mean it depends on $x_i$?
– TheLast Cipher
Jul 19 at 7:20
@TheLastCipher Yes.
– Salahamam_ Fatima
Jul 19 at 8:26
@TheLastCipher Yes.
– Salahamam_ Fatima
Jul 19 at 8:26
add a comment |Â
up vote
2
down vote
No, it is not correct. You start talking about $y$ without telling what $y$ is. But, above all, your approach only tells us why you think that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. The best approach, in this case, consists in providing a counterexample.
answered Jul 18 at 14:09
José Carlos Santos
114k1698177
I don't understand why arguing that something is not necessarily true is insufficient proof that it is not always true :? I have a hunch that this follow up falls under logic and philosophy? Does it?
– TheLast Cipher
Jul 18 at 14:34
What you said in your post was that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. That doesn't prove that there are cases in which it isn't a Cauchy sequence, which is what you were supposed to prove.
– José Carlos Santos
Jul 18 at 14:36
add a comment |Â
up vote
2
down vote
No, it is not correct. You start talking about $y$ without telling what $y$ is. But, above all, your approach only tells us why you think that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. The best approach, in this case, consists in providing a counterexample.
answered Jul 18 at 14:09
José Carlos Santos
114k1698177
I don't understand why arguing that something is not necessarily true is insufficient proof that it is not always true :? I have a hunch that this follow up falls under logic and philosophy? Does it?
– TheLast Cipher
Jul 18 at 14:34
What you said in your post was that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. That doesn't prove that there are cases in which it isn't a Cauchy sequence, which is what you were supposed to prove.
– José Carlos Santos
Jul 18 at 14:36
add a comment |Â
up vote
2
down vote
up vote
2
down vote
No, it is not correct. You start talking about $y$ without telling what $y$ is. But, above all, your approach only tells us why you think that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. The best approach, in this case, consists in providing a counterexample.
answered Jul 18 at 14:09
José Carlos Santos
114k1698177
No, it is not correct. You start talking about $y$ without telling what $y$ is. But, above all, your approach only tells us why you think that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. The best approach, in this case, consists in providing a counterexample.
answered Jul 18 at 14:09
José Carlos Santos
114k1698177
answered Jul 18 at 14:09
José Carlos Santos
114k1698177
answered Jul 18 at 14:09
José Carlos Santos
114k1698177
answered Jul 18 at 14:09
José Carlos Santos
114k1698177
114k1698177
I don't understand why arguing that something is not necessarily true is insufficient proof that it is not always true :? I have a hunch that this follow up falls under logic and philosophy? Does it?
– TheLast Cipher
Jul 18 at 14:34
What you said in your post was that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. That doesn't prove that there are cases in which it isn't a Cauchy sequence, which is what you were supposed to prove.
– José Carlos Santos
Jul 18 at 14:36
add a comment |Â
I don't understand why arguing that something is not necessarily true is insufficient proof that it is not always true :? I have a hunch that this follow up falls under logic and philosophy? Does it?
– TheLast Cipher
Jul 18 at 14:34
What you said in your post was that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. That doesn't prove that there are cases in which it isn't a Cauchy sequence, which is what you were supposed to prove.
– José Carlos Santos
Jul 18 at 14:36
I don't understand why arguing that something is not necessarily true is insufficient proof that it is not always true :? I have a hunch that this follow up falls under logic and philosophy? Does it?
– TheLast Cipher
Jul 18 at 14:34
I don't understand why arguing that something is not necessarily true is insufficient proof that it is not always true :? I have a hunch that this follow up falls under logic and philosophy? Does it?
– TheLast Cipher
Jul 18 at 14:34
What you said in your post was that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. That doesn't prove that there are cases in which it isn't a Cauchy sequence, which is what you were supposed to prove.
– José Carlos Santos
Jul 18 at 14:36
What you said in your post was that the sequence $bigl(f(x_n)bigr)_ninmathbb N$ might not be a Cauchy sequence. That doesn't prove that there are cases in which it isn't a Cauchy sequence, which is what you were supposed to prove.
– José Carlos Santos
Jul 18 at 14:36
add a comment |Â
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1
How do you choose your $y$ for it to work? And it should rather be $sigma (f(x_i),f(y))<epsilon '$ right?
– Suzet
Jul 18 at 13:58
What do you mean? I am not trying to make it work because I'm arguing it would not always work.
– TheLast Cipher
Jul 18 at 14:00
oh right. typo! thanks
– TheLast Cipher
Jul 18 at 14:01
I've corrected it now.
– TheLast Cipher
Jul 18 at 14:01
3
Oh, actually the statement is untrue, I didn't get it. Well, the only way to show that it is false is to give a counter example for it. Trying to explain why a proof of it wouldn't work is good, but you do not get a definitive conclusion before you give a counter example.
– Suzet
Jul 18 at 14:03