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Let $(a_{n})_{n=m}^{\infty}$ be a sequence of real numbers, and let $L$ be a real number. Then the following two statements are logically equivalent

(a) The sequence converges to $L$

(b) Every subsequence of $(a_{n})_{n=m}^{\infty}$ converges to $L$.

MY ATTEMPT

(a) Let us prove $(\Rightarrow)$ first.

Let $b_{n}$ be a subsequence of $a_{n}$. Thus there exists an strictly increasing function $f:\textbf{N}\rightarrow\textbf{N}$ such that $b_{n} = a_{f(n)}$.

Since $a_{n}$ converges to $L$, for every $\varepsilon > 0$, there is a natural number $N\geq m$ such that \begin{align*} n \geq N \Longrightarrow |a_{n} - L| \leq \varepsilon \end{align*}

But, since $f(n)$ is increasing, one has that $f(n) > n$.

Consequently, for every $\varepsilon > 0$, there exists a natural number $N\geq m$ such that \begin{align*} f(n) > n \geq N \Longrightarrow |a_{f(n)} - L| = |b_{n} - L| \leq \varepsilon \end{align*}

and $b_{n}\to L$ and $n\to\infty$, just as desired.

(b) We may now prove $(\Leftarrow)$.

Since every subsequence of $a_{n}$ converges to $L$ we can consider two particular cases.

More precisely, $b_{n} = a_{2n-1} \to L$ and $c_{n} = a_{2n} \to L$.

Consequently, since $b_{n}\to L$, for every $\varepsilon > 0$, there exists $N_{1} \geq 1$ such that \begin{align*} n \geq N_{1} \Longrightarrow |b_{n} - L| = |a_{2n-1} - L| \leq \varepsilon \end{align*}

Similarly, since $c_{n}\to L$, for every $\varepsilon > 0$, there exists $N_{2} \geq 1$, such that \begin{align*} n \geq N_{2} \Longrightarrow |c_{n} - L| = |a_{2n} - L| \leq \varepsilon \end{align*}

Finally, we conclude that, for every $\varepsilon > 0$, there exists a natural number $N = \max\{N_{1},N_{2}\}$ such that \begin{align*} n \geq N \Longrightarrow |a_{n} - L| \leq \varepsilon \end{align*}

Could someone please confirm if I am reasoning right?

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Yes, it is correct. But the proof of $\Leftarrow$ doesn't have to be so complex. Just use the fact that the whole sequence is a subsequence of itself.

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  • $\begingroup$ Thanks for the comment, José. I didn't realize it before. $\endgroup$ – BrickByBrick Apr 19 at 22:44

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