# Existence of countable approximative unit in $C_0(X) \iff X$ is $\sigma$-compact

Let $X$ be a locally-compact Hausdorff space. The following are equivalent:

1. There is a countable approximate unit $(u_n)_{n\in\mathbb{N}}$ for $C_0(X)$ such that $\|u_n\|\le 1$ for all $n\in\mathbb{N}$ and $(u_n)$ is monotone increasing

2. $X$ is $\sigma$-compact, that means: there is a sequence $(K_n)$ of compact subsets $K_n\subseteq X$ such that $X=\bigcup\limits_{n\in\mathbb{N}}K_n$.

My only idea was an idea for the direction $(2)\implies (1)$. If $C_0(X)$ is separable under this requirements, then a theorem from lecture give me (1). But I think the requirements are not enough to get separability of $C_0(X)$ (or follows, that X is metrizable?).

Other fact that I know is that every $C^*$-algebra has an approximate unit and I know the proof, but it doesn't help me I think... Do you know why (1) and (2) are equivalent or could you help me? Regards.

For $1) \implies 2)$, note that for an $u\in C_0(X)$ and any $\varepsilon > 0$, the set $\{x\in X : u(x) \geqslant \varepsilon \}$ is compact.

Let $K_n = \left\{x \in X : u_n(x) \geqslant \tfrac{1}{n+2}\right\}$. Since $u_n\in C_0(X)$, $K_n$ is compact. Furthermore, we have $K_n \subset \overset{\Large\circ}{K}_{n+1}$ since $(u_n)$ is monotonically increasing, whence $$K_n \subset \left\{x\in X : u_n(x) > \tfrac{1}{n+3}\right\} \subset K_{n+1},$$ and the middle set is open. Lastly, since $(u_n)$ is an approximative unit, we have $u_n(x) \to 1$ for all $x\in X$, so $X = \bigcup\limits_{n=0}^\infty K_n$, which shows $X$ is $\sigma$-compact.

For $2)\implies 1)$, note that since $X$ is locally compact and Hausdorff, you can assume that for all $n$ you have $K_n \subset \overset{\Large\circ}{K}_{n+1}$, and think of Urysohn's lemma.

We first show that we can make that assumption with the aid of the

Lemma: Let $Y$ be a locally compact Hausdorff space, $K \subset Y$ compact, and $U$ an open set containing $K$. Then there is an open $V$ with compact closure such that $K\subset V \subset \overline{V} \subset U$.

Proof: For every $y\in K$, there is a compact neighbourhood $K_y$ of $y$ contained in $U$. Let $V_y = \overset{\Large\circ}{K}_y$. Then $\{ V_y : y\in K\}$ is an open cover of $K$. Since $K$ is compact, there is a finite subset $F\subset K$ such that $K \subset V := \bigcup\limits_{y\in F} V_y$. Then $K \subset V$, and $$\overline{V} = \bigcup_{y\in F} \overline{V_y} \subset \bigcup_{y\in F} K_y \subset U$$ also shows that $\overline{V}$ is compact.

Then if we have $X = \bigcup\limits_{n=0}^\infty K_n$, we can construct a sequence $(M_n)$ of compact sets with $K_n \subset M_n \subset \overset{\Large\circ}{M}_{n+1}$ by applying the lemma with $U = X$ at each step. Let $M_0 = K_0$. If $M_n$ is already constructed, let $W_{n+1}$ be an open set with compact closure with $M_n \cup K_{n+1} \subset W_{n+1}$ as guaranteed by the lemma, and set $M_{n+1} = \overline{W_{n+1}}$. Then by construction $M_n \subset W_{n+1} \subset \overset{\Large\circ}{K}_{n+1}$ for all $n$, and also $K_n \subset M_n$ for all $n$, which shows $X = \bigcup\limits_{n = 0}^\infty M_n$.

Having shown that we may, we now assume that for all $n$ we have $K_n \subset \overset{\Large\circ}{K}_{n+1}$. Such a sequence of compact sets is called a normal exhaustion of $X$, and they are frequently used for various constructions on locally compact $\sigma$-compact Hausdorff spaces.

Compact spaces are normal, hence Urysohn's lemma guarantees the existence of a continuous $v_n \colon K_{n+2} \to[0,1]$ with $K_n \subset v_n^{-1}(\{1\})$ and $K_{n+2} \setminus \overset{\Large\circ}{K}_{n+1} \subset v_n^{-1}(\{0\})$. Then

$$u_n(x) = \begin{cases} v_n(x) &, x \in K_{n+2} \\ 0 &, x \notin K_{n+2}\end{cases}$$

is continuous on all of $X$, since the restrictions to the two open sets $\overset{\Large\circ}{K}_{n+2}$ and $X\setminus K_{n+1}$ which cover $X$ are continuous. Also, $(u_n)$ is an increasing sequence in $C_c(X) \subset C_0(X)$ converging pointwise to $1$. Since $\bigl(\overset{\Large\circ}{K}_n\bigr)$ is an increasing sequence of open sets exhausting $X$, every compact $K\subset X$ is contained in $K_n$ for all large enough $n$, which shows that $(u_n)$ is an approximative unit, as $$\lVert f - u_n\cdot f\rVert_\infty \leqslant \sup \{ \lvert f(x)\rvert : x \notin K_n\} \to 0.$$

• thank you. I needed a few hours to understand why I can assume $K_n\subset int(K_{n+1})$ (and i am a littlebit tired), thank you. But how to apply Urysohn's lemma? terrytao.wordpress.com/2009/03/02/… (lemma 1), have a such a function for every n and get a sequence? I'm not sure how to apply the lemma correctly – user151465 Nov 17 '14 at 4:07
• $K_{n+2}$ is compact, hence normal. Hence by Urysohn's lemma, there is a continuous function $f_n \colon K_{n+2} \to [0,1]$ with $f_n\lvert_{K_n} \equiv 1$ and $f_n\lvert_{K_{n+2}\setminus \overset{\circ}{K}_{n+1}} \equiv 0$. You can extend that to a continuous function $u_n$ on $X$ by setting $u_n(x) = 0$ for $x\notin K_{n+2}$ and $u_n(x) = f_n(x)$ for $x\in K_{n+2}$. That gives you the desired approximative unit, $0 \leqslant u_n \leqslant u_{n+1}$ and $u_n \nearrow 1$ (locally uniformly). – Daniel Fischer Nov 17 '14 at 10:42
• oh sometimes i am slow on the uptake.. thank you, i understand it. – user151465 Nov 18 '14 at 20:16
• @ms.mop We set $M_1 = \overline{W_1}$. Next, we apply the lemma to $K = M_1 \cup K_2$ and $U = X$. We get an open $W_2$ with compact closure and $M_1 \cup K_2 \subset W_2$. Then we set $M_2 = \overline{W_2}$. Continuing, if $M_n$ is already constructed, apply the lemma to $M_n \cup K_{n+1}$ to get an open $W_{n+1}$ with compact closure and $M_n \cup K_{n+1} \subset W_{n+1}$, set $M_{n+1} = \overline{W_{n+1}}$. Then we have $M_n \subset M_n \cup K_{n+1} \subset W_{n+1} \subset \overset{\Large\circ}{M}_{n+1}$, and $K_n \subset M_n$ for all $n$ gives $\bigcup M_n = X$. – Daniel Fischer Sep 23 '15 at 11:49
• That's great @ms.mop. And thanks for bringing the unclarity to my attention. – Daniel Fischer Sep 23 '15 at 13:24