Is it true that $(A_n)_n \to A \implies (A_{k_n})_n \to A$?

Question

Is it true that $(A_n)_n \to A \implies (A_{k_n})_n \to A$ in the set theoretic sense, and where $k: \mathbb{N} \to \mathbb{N}$ is a strictly increasing function.

I was able to show that it was true in the case of monotonic sequences, but am unsure about the general case.

Answer 1

Yes.

If $k_n\in\mathbb N$ with $k_1<k_2<k_3\dots$ and $B_n=A_{k_n}$ then: $$\liminf A_n\subseteq\liminf B_n\subseteq\limsup B_n\subseteq\limsup A_n\tag1$$

$\lim A_n$ exists iff $\liminf A_n=\limsup A_n$ and in that case we have: $$\liminf A_n=\lim A_n=\limsup A_n$$

The according to $(1)$ we also have $\liminf B_n=\limsup B_n=\lim A_n$ so that $\lim B_n$ exists with $\lim B_n=\lim A_n$.

Answer 2

Yes it is true since

$$\liminf A_n \subseteq \liminf A_{k_n} \subseteq \limsup A_{k_{n}} \subseteq \limsup A_n$$

Where $k_1 <k_2 <....<k_n<k_{n+1}<...$

Answer 3

Translating to indicator functions $$ A_n \to A \iff I(A_n)\to I(A) \tag{1} $$ where $I(A)$ is the indicator function of $A$. Thus if $A_n\to A$ and hence $I(A_n)\to I(A)$, it follows that $I(A_{n(k)})\to A$ by the corresponding result for sequences of real numbers. It follows that $$A_{n(k)}\to A. $$ To see that (1) is true note that $$ \begin{align*} I(A_n)\to I(A)&\iff\sup_{k\geq1}\inf_{n\ge k} I(A_n)=\inf_{k\geq1}\sup_{n\geq k} I(A_n)\\ &\iff \sup_{k\geq1}I(\cap_{n\geq k} A_{n})=\inf_{k\ge1}I(\cup_{n\geq k A_n})\\ &\iff I(\cup_{k\ge1}\cap_{n\geq k} A_{n})=I(\cap_{k\ge1}\cup_{n\geq k} A_n)\\ &\iff\liminf A_n=\limsup A_n \end{align*} $$