infimum of a closed set?

Question

Im trying to understand the proof of theorem $2.5$ part $2)$ of Daniel Lackers paper : A Strong form of Propagation of Chaos.

In essence :

$\textbf{Assume the following :}$ we have a metric space $X$ and a lower-semi-continuous non-negative function $H:X\to \mathbb{R}$ ($\{x : H(x)\leq c\}$ is compact). Also $H(x)=0 ~\iff~ x=y$.

$\textbf{The question is :}$ why for any neighborhood $U$ of $y$ we have,

$$\inf_{x\notin U}H(x)>0. $$

Im sure we need to use Any lower semicontinuous function $f: X \to \mathbb{R}$ on a compact set $K \subseteq X$ attains a min on $K$. ?

Sorry $H$ was non-negative i forgot to mention – Monty Dec 08 '19 at 17:20 — Monty, Dec 08 '19 at 17:20

score 1 · Accepted Answer · answered Dec 08 '19 at 17:24

1

The set $K=\{x:H(x)\leq 1\}$ is compact and thus so is $K\setminus U$. Thus as long as $K\setminus U$ is nonempty, $H$ achieves a minimum on $K\setminus U$, which is then also the minimum (and hence infimum) of $H$ on $X\setminus U$. Since $y\not\in K\setminus U$, this minimum must be positive. If $K\setminus U$ is empty, then $H(x)>1$ for all $x\not\in U$ and so the infimum of $H$ on $X\setminus U$ must be at least $1$.

answered Dec 08 '19 at 17:24

Eric Wofsey

342,377

Yes, yes makes sense, $K\backslash U=K\cap (X\backslash U)$ (where $X$ is the whole space) is compact since a compact set intersected with a closed set is compact, in a metric space. Thanks! – Monty Dec 11 '19 at 19:29

infimum of a closed set?

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