Timeline for Given locally compact and $\sigma$-compact, can we get partition of unity?
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6 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2015 at 7:17 | comment | added | Mathieu Baillif | 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. | |
| Sep 16, 2015 at 2:56 | comment | added | mafan | @MathieuBaillif:Why can $X$ be the union of the countable, increasing compact subsets? | |
| Sep 15, 2015 at 12:04 | comment | added | Mathieu Baillif | 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 $x\in K_n$. | |
| Sep 14, 2015 at 11:37 | comment | added | Ramiro de la Vega | Of course "$x$ belongs to finite number of $U_i$" may not hold. Let $X=\mathbb{R}$ and $K_i=[-i,i]$. | |
| Sep 14, 2015 at 9:56 | comment | added | mafan | @ArthurFischer:So I delete the post on Mathematics. | |
| Sep 14, 2015 at 8:45 | history | asked | mafan | CC BY-SA 3.0 |