Theorem. The following are equivalent:
- f is upper semi-continuous;
- for any sequence (x_{n}) in X (with possibly uncountable indices n) such that f(x_{n}) \geq f(x), if x_{n} \rightarrow x in X, then f(x_{n}) \rightarrow f(x) in \mathbb{R}.
Proof. Let f be upper semi-continuous. Given a sequence (x_{n}) in X such that f(x_{n}) \geq f(x), if x_{n} \rightarrow x in X, fix any \epsilon > 0. Then f^{-1}((-\infty, f(x) + \epsilon)) \subset X is open by upper semi-continuity of f and since it contains x, knowing that x_{n} \rightarrow x in X, we may find some x_{n} \in f^{-1}((-\infty, f(x) + \epsilon)). This implies that f(x_{n}) \in (-\infty, f(x) + \epsilon) \subset \mathbb{R}. Since f(x_{n}) \geq f(x) > f(x) - \epsilon, we have f(x_{n}) \in (f(x) - \epsilon, f(x) + \epsilon) \subset \mathbb{R}. This shows that f(x_{n}) \rightarrow f(x) in \mathbb{R}.
Conversely, suppose the second condition above. Fix a \in \mathbb{R}, and we aim to show that f^{-1}((-\infty, a)) \subset X is open. Fix any x \in f^{-1}((-\infty, a)). We want to show that there is an open neighborhood U \ni x in X such that U \subset f^{-1}((-\infty, a)). We now work by contradiction and say that there is no such U. In other words, for any neighborhood U_{n} \ni x in X, there exists x_{n} \in U_{n} such that f(x_{n}) \geq a > f(x). However, using the second condition, this implies that f(x) = \lim_{n \rightarrow \infty}f(x_{n}) \geq a > f(x), a contradiction. This finishes the proof. \Box
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