config.yml and set future: false.]]>So review: suppose $x_0\in f^{-1}(O)$, then to show $f^{-1}(O)$ open, we need to find some $\delta_{x_0}>0$ such that
\[B_{\delta_{x_0}}(x_0)\subseteq f^{-1}(O).\]I want the definition to be defined from a more intuitive engineering perspective. We know $f$ is continuous if $f(x_n)$ goes to $f(x)$ as $x_n$ goes to $x$, so a first intuitive try to come up with the definition by staring at the question is sequential, assuming we do not know the epsilon-delta kind of language of the definition of l.s.c.
We guess
\[\liminf_{y\to x} f(y) \geq f(x) \iff \sup_{n\in \mathbb N}\inf_{m\geq n} f(y_m)\geq f(x)\]for all
\[y_m\to x\]as $m\to \infty$.
So, guessing the idea is just that for usual continuity,
\[x_n\approx x \implies f(x_n)\approx f(x).\]Now it is directional signed, with allowable large difference in output: as we get close enough in the domain to some $x$ target, the function output is not less than…
\[x_n \approx x \implies f(x_n) \gtrapprox f(x).\]Once $y$ is sufficiently close to $x$, $f(y)$ cannot remain $\epsilon$ or more below $f(x)$. So basically, it cannot be too much lower than $f(x)$:
\[\forall \epsilon>0, \exists \delta>0: |y-x|<\delta \implies f(y) > f(x) -\epsilon.\]So you see it is not
\[f(y) > f(x) + \epsilon \implies f(y) > f(x),\]because that requires too much, as if $f(y)$ must be above $f(x)$, as if there must be upward jumps. We only require the looser $f(y) > f(x) -\epsilon$, which allows
\[f(y)<f(x).\]]]>