# Let $f:X\rightarrow \Bbb{R}\,\cup\,\{+\infty\}$ be a map. Then, for $x_0\in X,\;f(x_0)\geq \sup\limits_{V\in U(x_0)}\inf\limits_{x\in V}f(x)$

Let $$f:X\rightarrow \Bbb{R}\,\cup\,\{+\infty\}$$ be a map where $$X$$ is a real normed space. Then, for an arbitrary $$x_0\in X,$$ I want to prove that the following always holds:

\begin{align}f(x_0)\geq \sup\limits_{V\in U(x_0)}\inf\limits_{x\in V}f(x) \end{align}

MY TRIAL

Let $$x_0\in X,$$ then $$\exists\,V\in\,U(x_0)$$ such that $$f(x_0)\geq f(x),\;\forall\;x\in V$$, where $$U(x_0)$$ is the set of all neighbourhoods of $$x_0$$. So,

\begin{align}f(x_0)\geq \inf\limits_{x\in V}f(x),\;\;\text{for some}\;\;V\in U(x_0), \end{align} which implies that \begin{align}f(x_0)\geq \sup\limits_{V\in U(x_0)}\inf\limits_{x\in V}f(x) \end{align}

Kindly check if I am correct or wrong. If it happens that I'm wrong, kindly give an alternative proof. Thanks

• You need to give us a hint - what is $U(x_0)$? – David C. Ullrich Nov 24 '18 at 13:32
• @David C. Ullrich: Sorry! $U(x_0)$ is the set of all neighbourhoods of $x_0$ – Omojola Micheal Nov 24 '18 at 13:34
• @Mike Oh, so $X$ is a topological space? But why should that matter when $f$ is not necessarily continuos? – Hagen von Eitzen Nov 24 '18 at 13:35
• @Hagen von Eitzen: Yes! Infact, $X$ is a real normed space. – Omojola Micheal Nov 24 '18 at 13:36

There is not the slightest reason why $$V\in U(x_0)$$ with $$f(x_0)\ge f(x),\forall x\in V$$ should exist.
However, if $$V\in U(x_0)$$ then $$x_0\in V$$ (here, I make a wild guess that $$U(x_0)$$ denotes some non-empty subset of the power set of $$X$$ and each of its elements contains $$x_0$$) and hence $$\inf_{x\in V} f(x)\le f(x_0)$$.