Setting: Let $Y$ be a metric space and let $a_{n,k}\in Y$ for all $n\in\mathbb{N}$ and $k\in\mathbb{N}$. Suppose $a_{n,k}\to a_{\bullet, k}$ uniforly as $n\to\infty$. Suppose the sequences $\{a_{n,k}\}\to A_n$ as $k\to \infty$ for each $n\in\mathbb{N}$, .
There are two questions in my mind:
- Is it possible that $a_{\bullet,k}$ diverge as $k\to\infty$?
- Is it possible that $A_n$ diverges while $a_{\bullet,k}$ converge?
I appreciate help in all forms.
The motivation for me to ask these questions is that I know the following three properties
- boundedness
- unboundedness
- continuous
are closed under uniform convergent. That is, the limit of a uniformly convergent sequence of bounded function must be bounded, and vice versa for the other two properties, but I cannot prove nor disprove that convergence of sequences in metric space is also closed under uniform convergent.
Also, I have already proved the following Theorem by myself, and I was trying to figure out if the hypotheses are all necessary. I figured that if the hypothesis ($A_n\to A$) in the following Theorem is implied by the other two, then the statement convergent sequences in metric space are closed under uniform convergent is true. Thanks for any help.
The answer to the second question is no. If $\lim_{k\to\infty}a_{\bullet,k}=A$ then $\lim_{n\to\infty}A_n$ exists and equals $A$.
Proof: Let $\epsilon>0$. Let $N$ be such that $\forall n\ge N,\forall k,d(a_{n,k},a_{\bullet,k})<\epsilon/2$, and $K$ be such that $\forall k\ge K,d(a_{\bullet,k},A)<\epsilon/2$. Then, $$\forall n\ge N,\forall k\ge K,d(a_{n,k},A)<\epsilon$$ hence $$\forall n\ge N,d(A_n,A)\le\epsilon.$$
The answer to the first question is yes. Take for instance $$Y=\Bbb R^*,\quad a_{n,k}=\frac1n+\frac1k.$$
However, if $Y$ is complete, the answer to the first question is no (i.e. the answer to the question in the title is yes), because if $(a_{n,k})_k$ converges uniformly as $n\to\infty$ then it is uniformly Cauchy, hence $(A_n)_n$ is Cauchy.