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Here is the problem.

I have tried to prove the above statement as follows. Please let me know if my proof is correct and if anything is wrong, please let me know where the proof went wrong.

My attempt:

$X=(z_{2k-1})$

$Y=(z_{2k})$

To prove: $Z$ is convergent $\iff$ $X$ and $Y$ are convergent with $lim (X)=lim (Y)$

First Part:

$Z$ is convergent $\implies$ $X$ and $Y$ are convergent with $lim (X)=lim (Y)$:

1. $\forall$ $\epsilon>0 $ $\exists$ $N$ such that for all $n \ge N$: $|z_n-z|\lt\epsilon$

$\because $$2k\gt2k-1\ge k$, choose $k\ge N$

$\therefore$ to satisfy 1., $X$ and $Y$ both must converge to $z$

2nd Part:

$Z$ is convergent $\Leftarrow $ $X$ and $Y$ are convergent with $lim (X)=lim (Y)$

Suppose Z is not convergent & $X,Y$ are convergent with $lim (X)=lim (Y)=z$

2. $\exists$ $\epsilon>0 $ such that $\forall$ $N$ : $|z_n-z|\ge\epsilon$ for some $n\ge N$

$\forall$ $\epsilon>0 $ $\exists$ $K$ such that for all $k\ge K$: $|z_{2k-1}-z|\lt\epsilon$

which contradicts with 2. when N=K. Hence by contradiction, 2nd part is also proved.

Why I have doubt with the above proof is: Suppose there is a sequence $X$ which has two convergent subsequences $X'$ and $Y'$ (both converge to different values). Now the statement Convergence of $X'\implies$ Convergence of $X$ can be proven using contradiction as in 2nd part above. But that is not right. Please let me know why this is so. Thanks in advance. :)

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  • $\begingroup$ A sequence converges to $l$ if and only if all of its subsequences converge to $l.$ $\endgroup$ – Sahiba Arora Mar 24 '20 at 21:26
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The proof of the $\Leftarrow$ implication is not right. You introduce an element $z$, but you don't tell us what it is.

Suppose that $l=\lim X=\lim Y$. Take $\varepsilon>0$. There are natural numbers $N_1$ and $N_2$ such that$$n\geqslant N_1\implies\lvert x_n-l\rvert<\varepsilon\text{ and }n\geqslant N_2\implies\lvert y_n-l\rvert<\varepsilon.$$Let $N=\max\{N_1,N_2\}$. Then, if $n\geqslant N$, $n\geqslant N_1$, and therefore $z_{2n-1}=x_n$; so $\lvert z_{2n-1}-l\rvert<\varepsilon$. And, if $n\geqslant N$, $n\geqslant N_2$, and therefore $z_{2n}=y_n$; so $\lvert z_{2n}-l\rvert<\varepsilon$. Conclusion: the distance from $z_{2N-1},z_{2N},z_{2N+1},\ldots$ to $l$ is smaller than $\varepsilon$. So, if $n\geqslant\left\lceil\frac N2\right\rceil+1$, $\lvert z_n-l\rvert<\varepsilon$.

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  • $\begingroup$ Thanks a lot @José Carlos Santos. $z$ is equal to lim X= lim Y. I forgot to put that. I'll update my question accordingly. But apart from that is rest everything OK? $\endgroup$ – Koro Mar 24 '20 at 21:46
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    $\begingroup$ No. There is no contradiction between$$(\exists\varepsilon>0)(\forall N\in\mathbb N)(\exists n\in\mathbb N):n\geqslant N\text{ and }\lvert z_n-z\rvert\geqslant\varepsilon$$and$$(\forall\varepsilon>0)(\exists K\in\mathbb N)(\forall n\in\mathbb N):k\geqslant K\implies\lvert z_{2k-1}-z\rvert<\varepsilon.$$ $\endgroup$ – José Carlos Santos Mar 24 '20 at 21:52

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