Show that a lower semi continuous function composed with a continuous function is lower semi continuous Let $f_1$ and $f_2$ be two functions from $\mathbb{R}\rightarrow\mathbb{R}$. Furthermore, let $f_1$ be lower semi continuous and $f_2$ be continuous. Show that $f_1\circ f_2$ is lower semi continuous.
Here is my attempt at a solution:
Recall that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is lower semi continuous (LSC) if the set $\{x\in\mathbb{R} : f(x) > a\}$ is open $\forall a\in\mathbb{R}$. So, let $a\in\mathbb{R}$ and consider the set $U_a = \{x\in\mathbb{R} : f_1\circ f_2(x) > a\}$. Then we can also characterize $U_a$ as the pre-image, $U_a = (f_2^{-1}\circ f_1^{-1})(]a,\infty[)$. Since $f_1$ is LSC, we have that $f_1^{-1}(]a,\infty[)$ must be an open set in $\mathbb{R}$. Then, since $f_2$ is assumed to be continuous, the pre-image of the open set $f_1^{-1}(]a,\infty[)$ under $f_2$ must also be open. Then, since $a$ was an arbitrary element in $\mathbb{R}$, we have $(f_2^{-1}\circ f_1^{-1})(]a,\infty[)$ is open $\forall a\in\mathbb{R}$. Thus $f_1\circ f_2$ is LSC.
In the text I am reading, however, it is not given that $f$ is LSC $\iff$ the set $\{x\in\mathbb{R} : f(x) > a\}$ is open $\forall a\in\mathbb{R}$, this was something I learned in a course last year. Instead, we are given that $f$ is LSC at a point $x$ $\iff$ for every net $(x_a)_{a\in A}$ in $\mathbb{R}$, $x_a\rightarrow x \implies f(x)\leq \lim\inf(x_a)$. I am quite unfamiliar with nets, can someone give me a hint on how to show that these definitions of LSC are equivalent?
 A: Let's also state a localised version of lower semicontinuity in terms of neighbourhoods:
Definition: Let $X$ a topological space, and $f \colon X \to \mathbb{R}$ a function. Then $f$ is lower semicontinuous at $x$ if for every $a < f(x)$ the set $f^{-1}(]a,+\infty[)$ is a neighbourhood of $x$.
Then we can show that a function $f \colon X \to \mathbb{R}$ satisfies "$f^{-1}(],a+\infty[)$ is open for every $a \in \mathbb{R}$" if and only if it is lower semicontinuous at $x$ (in the sense of the definition above) for every $x\in X$.
For, if $f^{-1}(]a,+\infty[)$ is open, and $a < f(x)$, then $f^{-1}(]a,+\infty[)$ is a neighbourhood of $x$. And conversely, if $f$ is LSC at every $x$, then $f^{-1}(]a,+\infty[)$ is a neighbourhood of every $x\in X$ with $f(x) > a$ - but that is exactly $f^{-1}(]a,+\infty[)$, which thus is open.
Then we need to see that the two pointwise definitions are equivalent. If $(x_{\alpha})$ is a net converging to $x$, and $a < f(x)$, then there is an $\alpha_0$ such that $x_{\alpha} \in f^{-1}(]a,+\infty[)$ for all $\alpha \geqslant \alpha_0$, since $f^{-1}(]a,+\infty[)$ is a neighbourhood of $x$. Hence $\liminf f(x_{\alpha}) \geqslant a$. Since that holds for every $a < f(x)$, it follows that $\liminf f(x_{\alpha}) \geqslant f(x)$.
Conversely, if $f$ is not lower semicontinuous at $x$ in the sense of the definition above, there is an $a < f(x)$ such that $f^{-1}(]a,+\infty[)$ is not a neighbourhood of $x$. Hence for every neighbourhood $U$ of $x$ we can choose a point $x_U \in U \setminus f^{-1}(]a,+\infty[)$. The family of neighbourhoods of $x$ is a directed set if (partially) ordered by reversed inclusion (i.e. $U \leqslant V \iff V \subset U$), and thus we have a net $(x_U)$ converging to $x$. But by construction, $f(x_U) \leqslant a$ for all $U$, so $\liminf f(x_U) \leqslant a$, so $f$ is also not LSC at $x$ in the sense of the net-definition.
