Proving that $X$ is a metric space If $X$ is the set of all sequences $(x_k)_k$ such that $sup_{k\in N}|x_k|$ exists, then with $$d(x,y)=sup_{k\in N} |x_k-y_k|$$ is $X$ a metric space?
So, for this I have to prove the three properties of $d$ hold? I'm having trouble proving these, especially the triangle inequality.
Also, a followup question to this. Is the Heine-Borel theorem satisfied in this space?
 A: Hint:
$$d(x,z)+d(z,y)=\sup_{k\in\Bbb N}|x_k-z_k|+\sup_{k\in\Bbb N}|z_k-y_k|\geq \sup_{k\in\Bbb N}\left(|x_k-z_k|+|z_k-y_k|\right)$$
Now, note that by  the triangle inequality, we have, $$|x_k-z_k|+|z_k-y_k|\geq |(x_k-z_k)+(z_k-y_k)|=|x_k-y_k|$$
Can you take it from here?
Hint #2:
If $(a_n)_{n\in\Bbb N}$ and $(b_n)_{n\in\Bbb N}$ be two sequences (with supremum) with $a_n\geq b_n~\forall~n\in\Bbb N$, then we have $$\sup_{k\in\Bbb N}a_k\geq\sup_{k\in\Bbb N} b_k$$
(The proof should be obvious)
A: $X$ is the space of all real bounded sequences, which can be equipped with the supremum norm, defined by $\Vert x\Vert=\sup\{\vert x_k\vert\;\,k\in\mathbb{N}\}$.
It is not difficult to see that $\Vert . \Vert$ is indeed a norm. Hence one defines a distance on $X$ by setting, for all $(x,y)\in X^2$ : $d(x,y)=\Vert x-y\Vert$.
For your followup question, $X$ is not finite dimensional, hence the Heine-Borel proposition doesn't hold (by Riesz theorem), but this can seen directly :
Consider de closed unit ball $B=\{x\in X;\,\Vert x\Vert=1\}$. Let $x^{(k)}\in B$ defined by $\forall n\in\mathbb{N},\,x^{k}_n=\delta_{k,n}$ (Kronecker symbol, equal to 1 if $k=n$ and $0$ otherwise).
For $k\neq \ell$, we have $\Vert x^{k}-x^{\ell}\Vert=1$, so that the sequence $(x^{k})_{k\ge0}$ doesn't have any convergent subsequence.
A: (I). For every $k$ we have $|x_k-z_k|\leq |x_k-y_k|+|y_k-z_k|.$ Therefore $$d(x,z)=\sup_k|x_k-z_k|\leq \sup_k (|x_k-y_k|+|y_k-z_k|)\leq$$ $$\leq  (\sup_k|x_k-y_k|)+(\sup_k|y_k-z_k|)\quad \text { (...See Note below...) }=$$ $$=d(x,y)+d(y,z).$$ Note: Let $A=\sup_k|x_k-y_k|$ and $B=\sup_k |y_k-z_k|.$ For every $k$ we have $A\geq |x_k-y_k|$ and $B\geq |y_k-z_k|,$ implying that $A+B$ is an upper bound for the set $S=\{|x_k-y_k|+|y_k-z_k|\}_k.$ Therefore $\sup S\leq A+B.$ 
That is, $\sup_k(|x_k-y_k|+|y_k|)\leq (\sup_k|x_k-y_k|)+(\sup_k|y_k-z_k|).$
(II). For $n\in \Bbb N$ let $y(n)=(x_{n,k})_k$ where $x_{n,n}=1$ and $x_{n,k}=0$ when $k\ne n.$ The set  $Y=\{y(n):n\in \Bbb N\}$ is closed and bounded. But $Y$ is not compact because the family  $W=\{B_d(y(n),1/2):n\in \Bbb N\}$ is an infinite open cover of $Y$ and no proper subset of $W$ is a cover of $Y.$
The space $X$ (commonly called $l_{\infty}$) is not finite-dimensional. In a vector space of dimension $D<\infty$ any linearly independent subset has at most $D$ members. But $Y$ is an infinite linearly independent subset of $l_{\infty}.$
