# Prove the equivalence: the sequence $(a_{n})_{n=m}^{\infty}$ converges to $L$ iff every subsequence of $(a_{n})_{n=m}^{\infty}$ converges to $L$.

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?

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.

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