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User Amit Modi
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Answer:

Sure. Let $f_1, f_2, \dots, f_n$ be a family of continuous functions on a set $X$. Let $f(x) = \inf_{i=1}^n f_i(x)$. We want to show that $f$ is upper semicontinuous.

For any $x \in X$ and $\epsilon > 0$, we want to show that the set

$$\{ y \in X : f(y) < f(x) + \epsilon \}$$

is open. This set is the same as

$$\{ y \in X : \inf_{i=1}^n f_i(y) < f(x) + \epsilon \}.$$

Since the functions $f_i$ are continuous, for each $i$, the set

$$\{ y \in X : f_i(y) < f(x) + \epsilon \}$$

is open. Therefore, the set $\{ y \in X : f(y) < f(x) + \epsilon \}$ is the union of open sets, and so it is open.

This shows that $f$ is upper semicontinuous.

User Shak Ham
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