About convergent subsequence.

Question

Let $f_n$ be sequence of functions defined by: $$ f_n = \begin{cases} \cos(nx) & x\geq 1/n \\ x & x<1/n \end{cases} $$ My question is that if this sequence have subsequences which are converging to every points then what's the expression of that?My second question is that is there any subsequence converging to $0$ at every points of domain? And the third question is what kind of sequence is this? Thanks! Ok, I want to add something here, these $f_n$ are defined on E={1/m : m∈ℕ}.

Answer 1

If these are your questions,

Q 1) "Does this sequence have subsequences which are converging at every point?"

Q2) "Is there a subsequence converging to $0$ at every point of this domain?"

Q3)"What kind of sequence is this?"

Then I would propose the following answers

Ans $1$. Yes, there are subsequences of $f_n$ converging at every point of the domain.

Proof :- If $x\le 0$, then

$f_n(x)=x \quad ,\forall n$ since $1/n\gt 0\ge x ,\forall n\in \mathbb{N}$

So $\displaystyle{\lim}_{n\to \infty}f_n(x)=x$.

Hence every subsequence of $f_n$ converges for $x\le 0$

If $x\gt 0$, then by Archimidean property, $\exists m\in \mathbb{N}$ such that $1/m \lt x$. So $\forall n\gt m$

$f_n(x)=\cos (nx)$

Now notice that $\{f_n(x)\}_{n\ge m}$ is a bounded sequence and hence has a convergent subsequence.

This proves the claim.

Ans $2$) No, this is not true.

Becuase for $x\lt 0$,

$\displaystyle{\lim}_{n\to \infty}f_n(x)=x\lt 0$.

Then how can $f_n$ have a subsequence converging to $0$ for $x\lt 0$.

Ans $3$) I don't really understand what do you mean by "kind of sequence" . If it is for monotone or bounded, then the answer is no as can be easily seen.

Hope this helps.