# Given locally compact and $\sigma$-compact, can we get partition of unity?

Let $X$ be a locally compact, $\sigma$-compact Polish (complete and separable metric) space. How to prove: "There is an increasing sequence of continuous cut-off functions with compact support, $0\leqslant f_n(x) \leqslant 1$ and $f_n(x) \to 1$ as $n\to \infty"$?

My proof: let $X=\bigcup_{i=1}^{\infty} K_i$, each $K_i$ is compact. Then we can find a pre-compact neighborhood $U_i \supset K_i$. Using Urysohn's lemma we can find function $g_i=1$ on $K$ and $g_i=0$ on $X \backslash{U_i}$. Let $$g(x)=g_1(x)+g_2(x)+...$$ If for any point $x\in X$, $x$ belongs to finite number of $U_i$, then this function is well defined. Note that $g(x)\neq 0$ at any point $x\in X$ since $X=\bigcup_{i=1}^{\infty}K_i$. Thus we can define $$f_1=g_1/g, f_2=(g_1+g_2)/g,...,$$which is what we need.

However, it seems to me that "$x$ belongs to finite number of $U_i$" may not hold. So how to prove this theorem?

And how to relax the assumption on $X$ to get this sequence of functions $f_n$?

• @ArthurFischer:So I delete the post on Mathematics. – mafan Sep 14 '15 at 9:56
• Of course "$x$ belongs to finite number of $U_i$" may not hold. Let $X=\mathbb{R}$ and $K_i=[-i,i]$. – Ramiro de la Vega Sep 14 '15 at 11:37
• Using local and $\sigma$ compactness, you can write $X=\cup_{n=1}^\infty K_n$ where each $K_n$ is compact and contained in the interior of $K_{n+1}$. Set $f_n(x)$ to be $0$ outside of $K_n$ and the max between $1$ and the distance from $x$ to the boundary of $K_n$ when $x\in K_n$. – Mathieu Baillif Sep 15 '15 at 12:04
• @MathieuBaillif:Why can $X$ be the union of the countable, increasing compact subsets? – mafan Sep 16 '15 at 2:56
• Since $X$ is $\sigma$-compact, $X=\cup_{i=1}^\infty C_i$ with $C_i$ compact. Define $K_n$ by induction: cover $K_{n-1}\bigcup\cup_{i=1}^n C_i$ by relatively compact open sets, extract a finite subcover, $K_n$ is the closure of its union. – Mathieu Baillif Sep 16 '15 at 7:17