Show that: for a semi-continuous function $g: \mathbb{R}^n \rightarrow \mathbb{R}$, the set $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ is closed I'm super rusty on my real analysis and I'm stuck on this easy question.
Show that for a semi-continuous function $ g: \mathbb{R}^n \rightarrow \mathbb{R}$, the set $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ is closed
I have found some proofs online though I am working with somewhat different definitions of semi-continuous functions and closed sets and am trying to see whether sticking to the definitions I know.
In particular: 
A closed set is a set such that all limits points are within the set.
A point $x \in S$ is a limit point of $S$ if $\forall \delta > 0, \exists y \in S$ such that $y \neq x$ and  $|x-y| < \delta$.
A function $g$ is lower semicontinuous at $x$ if $\forall \epsilon >0, \space \exists \delta > 0$ such that $g(x) - g(y) < \epsilon \space \forall y \in N_{\delta}(x)$ 
 A: We have to show that if $x$ is a limit point of $\{x \in \mathbb{R}^n: g(x) \leq 0\}$ then $x$ is in that set, i.e. $g(x) \leq 0$.
First let's write what being a limit point tells us about $x$. It says that there are points $y$ such that $g(y) \leq 0$ as close to $x$ as you want :
$$\forall \delta > 0 \exists y \in N_\delta(x), g(y) \leq 0$$
Then write what lower semicontinuity of $g$ at $x$ means about the values of $g$ near $x$ . It says that if $y$ is sufficiently close to $x$ then $g(y)$ is at least as high as $g(x)- \varepsilon$:
$$\forall \epsilon > 0, \exists \delta > 0, \forall y \in N_\delta(x), g(x)-g(y) < \varepsilon$$
You see how they seem made for each other ?
Suppose $g(x) > 0$, and plug $\epsilon = g(x)$ in the lower semicontinuity property.
You get :
$$\forall \delta > 0 \exists y \in N_\delta(x), g(y) \leq 0$$
$$\exists \delta > 0 \forall y \in N_\delta(x), g(y) > 0$$
These two are the negation of each other so this is a contradiction.
The semicontinuity tells you about some $\delta > 0$, if you give it to the limit point property you learn about some $y \in N_\delta(x)$, such that $0 < g(y) \leq 0$.
So $g(x) > 0$ is contradictory, therefore $g(x) \leq 0$.
