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$(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ have the same limit.

Two convergent sequences in a metric space.Correctness of Analysis argument with Cauchy sequencesProve $aX_n +bY_n$ is a Cauchy Sequence.If $(x_n)$ and $(y_n)$ are Cauchy sequences, then give a direct argument that $ (x_n + y_n)$ is a Cauchy sequenceIf $x_n$ and $y_n$ are Cauchy then $leftfrac2x_ny_nright$ is CauchyProve that if $x_n$ is Cauchy, eventually positive and its limit isn't $0$ there's a $delta > 0$ such that $x_n - delta$ is eventually positive.My Proof: Every convergent sequence is a Cauchy sequence.$x_n to x in mathbbR$ and $y_n to 0 $. What about $fracx_ny_n$?Show that in $mathbbR^n$, sum of two Cauchy sequences is Cauchy.Prove that $preccurlyeq$ is a linear ordering over the set of all equivalence classes of Cauchy sequences of rationals

0

$begingroup$

Assume that $f: mathbbR -0to mathbbR$ is uniformly continuous.
Assume $(x_n)_ninmathbbNin(mathbbR-0)^mathbbN$ and $(y_n)_ninmathbbNin(mathbbR-0)^mathbbN$ are both sequences that converge to zero. Show that $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ have the same limit.

To start, I proved that $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ are Cauchy sequences (which converge). But I'm not sure how to proceed. Heuristically, I want to show that the distance between $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ is very small.

My attempt:

Since $f$ is uniformly continuous, then for all $epsilon>0$ there exists $delta>0$ such that if $|x-y|<delta_epsilon$ then $|f(x)-f(y)|<epsilon.$

Let's say for all $eta>0$, if $n>N_1$ then $|x_n|<eta$ and if $n>N_2$ then $|y_n|<eta$.

Pick $N=max(N_1,N_2)$ such that when $n>N$ one has $$|x_n-y_n|leq |x_n|+|y_n|<eta/2+eta/2=eta.$$

But how do I make the leap to concluding something about $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$?

real-analysis cauchy-sequences

share|cite|improve this question

edited Mar 15 at 3:18

Math Enthusiast

asked Mar 15 at 3:06

Math Enthusiast Math Enthusiast

555

$endgroup$

add a comment | 

0

$begingroup$

Assume that $f: mathbbR -0to mathbbR$ is uniformly continuous.
Assume $(x_n)_ninmathbbNin(mathbbR-0)^mathbbN$ and $(y_n)_ninmathbbNin(mathbbR-0)^mathbbN$ are both sequences that converge to zero. Show that $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ have the same limit.

To start, I proved that $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ are Cauchy sequences (which converge). But I'm not sure how to proceed. Heuristically, I want to show that the distance between $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ is very small.

My attempt:

Since $f$ is uniformly continuous, then for all $epsilon>0$ there exists $delta>0$ such that if $|x-y|<delta_epsilon$ then $|f(x)-f(y)|<epsilon.$

Let's say for all $eta>0$, if $n>N_1$ then $|x_n|<eta$ and if $n>N_2$ then $|y_n|<eta$.

Pick $N=max(N_1,N_2)$ such that when $n>N$ one has $$|x_n-y_n|leq |x_n|+|y_n|<eta/2+eta/2=eta.$$

But how do I make the leap to concluding something about $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$?

real-analysis cauchy-sequences

share|cite|improve this question

edited Mar 15 at 3:18

Math Enthusiast

asked Mar 15 at 3:06

Math Enthusiast Math Enthusiast

555

$endgroup$

add a comment | 

$begingroup$

Assume that $f: mathbbR -0to mathbbR$ is uniformly continuous.
Assume $(x_n)_ninmathbbNin(mathbbR-0)^mathbbN$ and $(y_n)_ninmathbbNin(mathbbR-0)^mathbbN$ are both sequences that converge to zero. Show that $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ have the same limit.

To start, I proved that $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ are Cauchy sequences (which converge). But I'm not sure how to proceed. Heuristically, I want to show that the distance between $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$ is very small.

My attempt:

Since $f$ is uniformly continuous, then for all $epsilon>0$ there exists $delta>0$ such that if $|x-y|<delta_epsilon$ then $|f(x)-f(y)|<epsilon.$

Let's say for all $eta>0$, if $n>N_1$ then $|x_n|<eta$ and if $n>N_2$ then $|y_n|<eta$.

Pick $N=max(N_1,N_2)$ such that when $n>N$ one has $$|x_n-y_n|leq |x_n|+|y_n|<eta/2+eta/2=eta.$$

But how do I make the leap to concluding something about $(f(x))_ninmathbbN$ and $(f(y))_ninmathbbN$?

real-analysis cauchy-sequences

share|cite|improve this question

edited Mar 15 at 3:18

Math Enthusiast

asked Mar 15 at 3:06

Math Enthusiast Math Enthusiast

555

$endgroup$

share|cite|improve this question

edited Mar 15 at 3:18

Math Enthusiast

asked Mar 15 at 3:06

Math Enthusiast Math Enthusiast

555

share|cite|improve this question

edited Mar 15 at 3:18

Math Enthusiast

asked Mar 15 at 3:06

Math Enthusiast Math Enthusiast

555

asked Mar 15 at 3:06

Math Enthusiast Math Enthusiast

555

asked Mar 15 at 3:06

Math Enthusiast Math Enthusiast

555

1 Answer
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Your $eta$ is arbitrary. Choose it such that $eta = eta_epsilon < delta_epsilon$.

Then by the uniform continuity $|f(x_n)-f(y_n)|<epsilon$, for all $n>N$.

share|cite|improve this answer

answered Mar 15 at 3:29

user647486user647486

44918

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1

$begingroup$

Your $eta$ is arbitrary. Choose it such that $eta = eta_epsilon < delta_epsilon$.

Then by the uniform continuity $|f(x_n)-f(y_n)|<epsilon$, for all $n>N$.

share|cite|improve this answer

answered Mar 15 at 3:29

user647486user647486

44918

$endgroup$

add a comment | 

1

$begingroup$

Your $eta$ is arbitrary. Choose it such that $eta = eta_epsilon < delta_epsilon$.

Then by the uniform continuity $|f(x_n)-f(y_n)|<epsilon$, for all $n>N$.

share|cite|improve this answer

answered Mar 15 at 3:29

user647486user647486

44918

$endgroup$

add a comment | 

$begingroup$

Your $eta$ is arbitrary. Choose it such that $eta = eta_epsilon < delta_epsilon$.

Then by the uniform continuity $|f(x_n)-f(y_n)|<epsilon$, for all $n>N$.

share|cite|improve this answer

answered Mar 15 at 3:29

user647486user647486

44918

$endgroup$

share|cite|improve this answer

answered Mar 15 at 3:29

user647486user647486

44918

answered Mar 15 at 3:29

user647486user647486

44918

answered Mar 15 at 3:29

user647486user647486

44918