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I recall that, if $\psi:\Bbb{R}\longrightarrow \Bbb{R}$ is a function defined over $\Bbb R$ with euclidean topology, we have

$\liminf\limits_{y\to x} \psi(y) = \sup\limits_{U\in \mathscr{U}_x} \inf\limits_{y\in U\setminus \{x\}} g(y)$

where I called $\mathscr{U}_x$ the filter of neighborhoods at point $x$.

My goal is to prove that the function $f:\Bbb{R}\longrightarrow\Bbb{R}$ defined as

$f(x) = \liminf\limits_{y\to x} g(y)$

is lower semicontinuous, that is the set $L_f(t)=\{x\in \Bbb{R}\mid f(x)>t\}$ is open for every $t\in \Bbb{R}$. I noted that this is false if the topology over $\Bbb R$ is not Kolmogoroff or worse $T_1$, but that seems true for topologies with a decent grade of separation.

Cameron Buie
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Horatio
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    Try to unwind the definition. If $x \in L_f(t)$, then for some small $\epsilon > 0$ there is a neighborhood $U$ of $x$ such that $inf_{y \in U\backslash{x}} g(y) > t + \epsilon$. Now does $U$ lie in $L_f(t)$? –  Nov 16 '12 at 20:36
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    I assume $g$ and $\psi$ are supposed to be the same? – Nate Eldredge Nov 16 '12 at 21:10

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