Definition of upper semi-continuous functions: limsup or liminf?

Question

Notation: $\{f\geq c\}$ stands for $\{x\in x: fx\geq c\}$.

The standard definition of an upper semi-continuous function $f:X\to \bar{ \mathbb R}$ is:

1. For each $c$ in $\mathbb R, \{f\geq c\}$ is closed

Or equivalently

2. For each net $x_a\to x, \limsup_a fx_a\leq fx$

It seems to me that the $ \limsup_a$ here can be substituted by $\liminf_a$:

3. For each net $x_a\to x, \liminf_a fx_a\leq fx$

or even by:

4. For any $x_a\to x$ s.t. $fx_a$ converges, $\lim fx_a\leq fx$.

Of course 2) implies 3) and 4); FOR THE CONVERSE: Let $x_a$ be any net in $\{f\geq c\}$ with $x_a\to x$, we have $fx\geq \liminf_afx_a\geq c$, hence $\{f\geq c\}$ is closed.

Is this true?

Answer 1

I think all these conditions are equivalent. When taking $\limsup$/$\liminf$ we can extract a subnet that has the corresponding limit, so all definitions are equivalent.

The confusion may stem from the fact that upper-semicontinuity of $f$ at $x_0$ is defined by $$ \limsup_{x\to x_0} f(x) \le f(x_0) $$ using $\limsup$ of functions. Using $\liminf$ here would not yield an equivalent characterization.

Take $$ f(x) = \begin{cases} -1 & \text{ if } x<0\\ 0 & \text{ if } x=0\\ +1 & \text{ if } x>0\end{cases} .$$ Then $\liminf_{x\to0} f(x) = -1 < f(0) < \limsup_{x\to0}f(x)$.