Compactness in the metric space of all bounded and unbounded complex sequences

Question

Let s be the set of all (bounded and unbounded) complex sequences and the metric on s is $$d(x,y)=\sum\limits_{j=1}^{\infty} \frac{1}{2^j}\frac{|x_j-y_j|}{1+|x_j-y_j|}$$ where $x=(x_j),y=(y_j)$. Now suppose $M$ is an infinite subset of s. I have to show that the necessary and sufficient condition for compactness of $M$ is that there exists a sequence of positive real numbers $\gamma_1, \gamma_2,\ldots $ such that for all $x$ in $M$ $|x_k|\leq \gamma_k$.

Answer 1

If $M$ is compact, consider the projections $p_k : M \rightarrow \mathbb{C}$ defined by $p_k(x)=x_k$. These are continuous, so are bounded by, say $\gamma_k$.

On the other hand, the proposition that every $M \subset \prod_{j=1}^\infty B[0; \gamma_j] \subset \mathbb{C} ^ \mathbb{N}$ is compact, is false. Consider for example $M=\prod_{j=1}^\infty B(0; 1)$ and $((x_j^k)_j)_k$ where $(x_j^k)_j=(\frac{k-1}{k}) \in M$. Then $\lim x^k=(1, 1, 1, 1, ...) \not\in M$.