Equivalence for upper semi-continuity in a metric space

Question

I believe this question is not hard, but I simply can't understand properly how to deal with the definition of upper semi-continuity of $\phi$ at $x$ in a metric space $X$, i. e. $\forall \epsilon>0, \exists \delta > 0 , \forall y \in X, d(x,y) < \delta \Rightarrow \phi(y) < \phi(x) + \epsilon$.

Basically, I must show the equivalence

$\phi$ upper semi-continuous at $x \Leftrightarrow \phi(x) = \lim_{n\rightarrow \infty} [ \sup \phi(B_X(x,1/n))]$.

I've tried working with sequences in a sequence of balls with center $x$ and decreasing radius $1/n$, but I believe I may lack some structure to advance further.

Answer 1

Let $\epsilon>0$, wlog assume that $\phi$ is bounded on $B(x,1),$ and choose an integer $N$ such that $1/N<\delta$ where $\delta $ satisfies the definition of upper-semicontinuity. Then if $n>N$, we have $\phi(B_X(x,1/n))\subseteq (-\infty ,\phi(x)+\epsilon)\Rightarrow a_n:=\sup\left \{ \phi(y):y\in B(x,1/n) \right \}\in (-\infty, \phi(x)+\epsilon].$ Now observe that $(a_n)$ is decreasing and bounded, so $\lim a_n=a$ exists and $a\le \phi(x)\ $ (because $\epsilon>0$ is arbitrary). But $\phi(x)\in B(x,1/n)$ for each integer $n$ so $a_n\ge \phi(x)\Rightarrow a\ge \phi(x);\ $ that is, $a=\phi(x).$

The other way is easier: just note that if $\phi(x) = \lim_{n\rightarrow \infty} [ \sup \phi(B_X(x,1/n))],$ then $(a_n)$ where $a_n:=\sup \phi(B_X(x,1/n))$ is a decreasing sequence that converges to $\phi(x).$ With this fact, and the definition, the result follows.