Using the definition of a subsequence given by Tao,
Let $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ be sequences of real numbers. We say that$(b_n)_{n=0}^{\infty}$ is a subsequence of $(a_n)_{n=0}^{\infty}$ iff there exists a function $f :\mathbb{N}\to \mathbb{N}$ which is strictly increasing, that is, $f(n+1)>f(n)$ for all $n\in \mathbb{N}$ such that $b_n=a_{f(n)}$ for all $n\in\mathbb{N}$.
I want to prove the following: Proposition $6.6.5$. A sequence $(a_n)_{n=0}^{\infty}$ converges to $L$ $\Longleftrightarrow$ every subsequence of $(a_n)_{n=0}^{\infty}$ converges to $L$.
As an aside, $f$ must necessarily be injective. If it weren't, then we would have some $n\neq n'$ and $f(n)=f(n')$. Without loss of generality, let $n>n'$. Then $n=n'+k$ with $k\in\mathbb{N}-\{0\}$ so $f(n'+k)=f(n')$. This however, violates the strictly increasing requirement on $f$, so it must be injective. Also, any strictly increasing function is injective. For the sake of contradiction, let $f(n')=f(n)$ and let $f$ be a strictly increasing function. Suppose $n'<n$. Then we have $f(n')<f(n)$. But this is a contradiction. So $n\leq n'$. If $n<n'$. Then $f(n)<f(n')$, also a contradiction, so $n\geq n'$. This implies that, by anti-symmetry, $n=n'$.
I think that the $\Longleftarrow$ direction is quite straightforward:
Since $(a_n)_{n=0}^{\infty}$ is a subsequence of itself–the function $f(n):=n$ is a mapping that satisfies the conditions of being a strictly increasing function from $N\to N$ and $(b_n)_{n=0}^{\infty}=(a_n)_{n=0}^{\infty}$–justified by the fact that the property of being a subsequence is reflexive and transitive, then $(a_n)_{n=0}^{\infty}$ converges to $L$ when all subsequences of $(a_n)_{n=0}^{\infty}$ converge to $L$.
For the $\Longrightarrow$ direction, I am having some issues.
We have that $(a_n)_{n=0}^{\infty}$ converges to $L$. Let $(b_n)_{n=0}^{\infty}$ be an arbitrary subsequence of $(a_n)_{n=0}^{\infty}$ with a function $f :\mathbb{N}\to \mathbb{N}$ satisfying $f(n+1)>f(n)$ such that $(b_n)_{n=0}^{\infty}=(a_{f(n)})_{n=0}^{\infty}$.
My main question is if any such function $f(n)$ is always greater than or equal to $n$. Because if it is, then I can proceed by the following method:
For any $\varepsilon>0$ there exists an $N\in\mathbb{N}$ such that for any $(n\geq N)\in\mathbb{N}$, we have $|a_n-L|<\varepsilon$. Since $f(n)\in\mathbb{N}$ and $f(n)\geq n\geq N$, then $|b_n-L|=|a_{f(n)\geq n}-L|<\varepsilon \Longrightarrow (b_n)_{n=0}^{\infty}=L$. Since $(b_n)_{n=0}^{\infty}$ was arbitrary, this means that any subsequence of $(a_n)_{n=0}^{\infty}$ will converge to $L$ if $(a_n)_{n=0}^{\infty}$ converges to $L$.
I have tried thinking of counterexamples of a function $f : \mathbb{N}\to\mathbb{N}$ such that $f(n+1)>f(n)$ where $f$ is injective and $f(n)<n$, for example, $f(n):=\frac{n}{2}$. This functions domain is $\mathbb{N}$ and by restricting its range so that $\frac{n}{2}\in\mathbb{N}$, then we have $f(n)<n$ but $f(n+1)$ does not exist. However, $f(n+2)$ does exist and $f(n+2)>f(n)$, but this is not the same requirement as in the definition, or at least, if it is, then there is a subtlety I am missing like $f(n+k)>f(n)$. Furthermore, this function would generate the same sequence by only taking even inputs. I guess this is more a question about if every $n\in\mathbb{N}$ must have an $f(n)\in\mathbb{N}$ for this proof to work. Also, any hints, comments, or better proofs/solutions would be appreciated. Thanks.
Best Answer
It follows from induction. For $n=0$ we have $f(0) \in \mathbb{N} \implies f(0) \geq 0 $ . Suppose $f(n) \geq n$ for some $n \geq 0$. Now notice that $f(n+1) > f(n)$. So $f(n+1) \geq f(n)+1 \geq n+1$.