$l_\infty $ and Stone-Cech compactification of $\mathbb{N}$

Question

$l_\infty$ is the vector space of real bounded sequences with the norm $$d(x,y)=\sup\{|x_n-y_n|, n\in \mathbb{N}\}.$$ I need to show that there is an isomorphism $T$ between $C(\beta \mathbb{N})$ and $l_\infty$ such that: $$d(T(x),T(y))=d(x,y)$$

I am a bit lost here. I do know that if $Y$ is a compact Hausdorff space, for a specified continuous $f$, there is a continuous $c_f:\beta \mathbb{N}\rightarrow Y $ such that we can make this diagram commute:

$\require{AMScd}$

\begin{CD} \mathbb{N} @>{\epsilon_\mathbb{N}}>> \beta\mathbb{N}\subset \prod_{f\in C_b(\mathbb{N})}I_f\\ @V{f}VV \\ Y \end{CD}

I suppouse we could take $T(c_f)=(f(n))_{n\in\mathbb{N}}$ if Y is compact in $\mathbb{R}$, but I have serious doubts about how to use this.

Answer 1

There is a problem in that you have not actually specified a $T:C(\beta\Bbb N)\to\ell_\infty$. What is the space $Y$?

Denote by $\iota$ the canonical inclusion $\Bbb N\hookrightarrow\beta\Bbb N$. Any $f\in C(\beta\Bbb N)$ restricts to a continuous $f\circ\iota:\Bbb N\to\Bbb R$, i.e. a sequence of real numbers. This is the natural thing to consider, and it's on us to check that:

• $(\rm i)$ all such sequences represented by $f\iota$ are bounded (then we can define $T:f\mapsto(f\iota(n))_{n\in\Bbb N}$ as a valid function $C(\beta\Bbb N)\to\ell_\infty$)
• $\rm(ii)$ to every $\alpha\in\ell_\infty$ there is a unique extension $f\in C(\beta\Bbb N)$ with $Tf=\alpha$
• $\rm(iii)$ the map $T$ is bicontinuous

$\rm(i)$ is not too bad. $\beta\Bbb N$ is compact, so for $f\in C(\beta\Bbb N)$ we know $f$ has bounded image - which specialises to the claim.

Let's settle $(\rm ii)$. If $\alpha\in\ell_\infty$ then there is some bound $K>0$ with $\alpha_n\in[-K,K]$ for all $n\in\Bbb N$. $[-K,K]$ is a compact Hausdorff space, and we have a continuous $\tilde{\alpha}:\Bbb N\to[-K,K]$ representing our sequence.

... so by the defining property of $\beta\Bbb N$ we have a unique extension $\tilde{f}:\beta\Bbb N\to[-K,K]$ of $\tilde{\alpha}$ along $\iota$. Defining $f$ to be the composite $\beta\Bbb N\to[-K,K]\hookrightarrow\Bbb R$ we find $f\in C(\beta\Bbb N)$ with $Tf=\alpha$ and because $\tilde{f}$ is unique, $f$ is unique.

$\rm(iii)$ is a little harder. We will show $T$ is an isometry, and $\rm(iii)$ follows. Notice that compactness implies $\|f-g\|_\infty$ is always attained at some $x\in\beta\Bbb N$, for two given $f,g\in C(\beta\Bbb N)$. Remember that every neighbourhood of $x$ contains points from $\iota(\Bbb N)$ - the inclusion is dense. Try to use this - and continuity of $f,g$ - to show $\|f-g\|_\infty\le\|Tf-Tg\|_\infty$.

Take a neighbourhood $U$ of $f(x)$ and a neighbourhood $V$ of $g(x)$. $f^{-1}(U)\cap g^{-1}(V)$ some (infinitely many) $\iota(n)$, $n\in\Bbb N$ since it is nonempty and $\iota(\Bbb N)$ is dense. Fixing $\epsilon>0$, we can choose $U=(f(x)-\epsilon/2,f(x)+\epsilon/2)$ and $V=(g(x)-\epsilon/2,g(x)+\epsilon/2)$. We find (infinitely many) naturals $m$ (near to $x$) with $|f(m)-f(x)|<\epsilon/2$ and $|g(m)-g(x)|<\epsilon/2$. It follows that $||f(m)-g(m)|-|f(x)-g(x)||<\epsilon$ by the triangle inequalities. As $\epsilon$ is arbitrary, we find (why?) $\sup_{m\in\Bbb N}|f(m)-g(m)|\ge|f(x)-g(x)|$, or in other words, $\|Tf-Tg\|_\infty\ge\|f-g\|_\infty$. Moreover, $\|f-g\|_\infty\ge\|Tf-Tg\|_\infty$ is immediate (why?) so we get to conclude $T$ is an isometry.

(if anyone is wondering why I wrote this after Alex's answer, it was mostly so that I could do the exercise as well!)

Answer 2

You've only stated one half of the universal property of the Stone-Cech compactification. If $Y$ is compact Hausdorff, then every continuous map $X\to Y$ extends uniquely to a continuous map $\beta X\to Y$. In other words, the continuous maps $X\to Y$ are in bijection with the continuous maps $\beta X\to Y$, via composition with the canonical map $X\to \beta X$.

Now in the case of the discrete space $\mathbb{N}$, a continuous function $\mathbb{N}\to Y$ is a sequence from $Y$, so the sequences from $Y$ are in bijection with the continuous maps $\beta\mathbb{N}\to Y$.

We can't apply this observation directly to your question, since (a) $C(\beta\mathbb{N})$ is the continuous functions to $\mathbb{R}$, which is not compact, and (b) $\ell^\infty$ is the space of bounded sequences, not all sequences.

But note that since $\beta\mathbb{N}$ is compact, its image along any continuous map is compact, and hence bounded. And on the other side, a bounded sequence from $\mathbb{R}$ is the same thing as a sequence from a bounded subset of $\mathbb{R}$.

So we have the chain of bijections (using the fact that $[-n,n]$ is compact for all $n$): \begin{align*} C(\beta\mathbb{N}) &\cong \bigcup_{n\in \mathbb{N}} \mathrm{Hom}(\beta\mathbb{N},[-n,n])\\ &\cong \bigcup_{n\in \mathbb{N}} \mathrm{Hom}(\mathbb{N},[-n,n])\\ &\cong \ell^\infty. \end{align*}

I'll leave it to you to check that this is an isomorphism of normed vector spaces.

Answer 3

We can define $\beta\mathbb{N}$ as the unique compact Hausdorff space in which $\mathbb{N}$ is dense and $C^*$-embedded, i.e., any bounded continuous function from $\mathbb{N}$ to $\mathbb{R}$ (thus any element of $\ell^\infty$) extends to a continuous function from $\beta\mathbb{N}$ to $\mathbb{R}$. In other words, any element $f\in l^\infty(\mathbb{N})$ extends to (unique from density) $f^\beta\in C(\beta\mathbb{N})$, and moreover, this function $f\mapsto f^\beta$ is clearly a ring isomorphism (the inverse function being restriction i.e. $f^\beta\restriction_\mathbb{N} = f$). For example, $(f+g)^\beta = f^\beta+g^\beta$ since both of these functions agree on $\mathbb{N}$.

One can show that for isomorphisms between rings of bounded continuous functions, say $C^*(X)$ and $C^*(Y)$, the sup norm is always preserved.

Answer 4

Every Tychonoff space $X$ is a subspace of its Stone-Čech compactification $\beta X,$ and the restriction map $$T:C(\beta X)\to C_b(X)$$ is an isomorphism of algebras, where $C$ means "continuous functions" (with real or complex values) and $\cdot_b$ means "bounded". Moreover, since $X$ is dense in $\beta X,$ $T$ preserves the $\sup$ norm (because for every continuous function $g$ on a space $Y\supset X,$ $g(\overline X)\subset\overline{g(X)}$: apply this to $Y=\beta X$ and $g=|f|$ for $f\in C(Y)$).

Applying this to $X=\Bbb N$ (with the discrete topology) immediately gives an isomorphism of normed algebras $$T:C(\beta\Bbb N)\to C_b(\Bbb N)=\ell^\infty.$$