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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?

user23899
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  • This space is called $l_{\infty}$, pronounced el-infinity. – DanielWainfleet Oct 04 '17 at 05:58
  • @DanielWainfleet What does the heine-borel theorem tell us about this? – user23899 Oct 04 '17 at 06:26
  • The Heine-Borel theorem (generalized from dimension $1$) is that with the (usual) Cartesian metric on $\Bbb R^n$ with $n\in \Bbb N,$ a set is compact iff it is closed and bounded. It does not apply to $X.$ It will tell us that if $S$ is a finite-dimensional subspace of $X$ and if $T\subset S$ then $T$ is compact iff $T$ is bounded, and closed in $X$. – DanielWainfleet Oct 04 '17 at 06:57

3 Answers3

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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)

Prasun Biswas
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  • I don't understand how you can remove the $sup$. Can you explain what you did there? – user23899 Oct 04 '17 at 04:24
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    I'm not removing the sup. I have left a few gaps intentionally in my post for you to fill. If $(a_n),(b_n)$ be two sequences such that $a_n\geq b_n$, then we have $\sup_k a_n\geq\sup_k b_n$. Also, noting that $\sup_{k\in\Bbb N}|x_k-y_k|=d(x,y)$, conclude your proof. – Prasun Biswas Oct 04 '17 at 04:27
  • Okay, I'll try to work off of this. Can you comment on my followup question? Essentially is it true that a set $K$ is compact in our metric space $X$ iff $K$ is bounded and closed? – user23899 Oct 04 '17 at 04:31
  • @user23899, I just noticed that my argument with the triangle inequality was a bit flawed. I just revised my answer, so perhaps it might help you understand better. We have applied $|a|+|b|\geq |a-b|$ which is a slight variant of the usual triangle inequality. – Prasun Biswas Oct 04 '17 at 04:34
  • I'm still not seeing how the proof is obvious from what you gave. I'm not able to connect the two hints. – user23899 Oct 04 '17 at 04:57
  • @user23899, $\sup$ is order-preserving, meaning that from $\alpha \geq \beta$, you can deduce $$\sup_{k \in \mathbb{N}} \alpha \geq \sup_{k \in \mathbb{N}} \beta.$$ So use the inequality on the second line of math in Prasun Biswas's answer to help you continue the reasoning in the first line of math. – goblin GONE Oct 04 '17 at 05:01
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$X$ 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.

Adren
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  • What do you mean by $X$ is not finite dimensional? I'm having trouble following how your explanation implies that the theorem doesn't hold – user23899 Oct 04 '17 at 05:15
  • The sequences $x^{(k)}$ defined in my previous answer are linearly independent. Hence $X$ is not a finite dimensional vector space. The theorem of F. Riesz tells us that the closed unit ball cannot be compact in such a space. – Adren Oct 04 '17 at 06:34
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(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 $l_{\infty}.$

DanielWainfleet
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