Closure set in $\mathbb{R}$ Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function.

Let $$A=\left\{x\in\mathbb{R}, f(x)&gt; 0\right\}$$

Can you please help me to find the closure set of $A$?

Thanks in advance.
for every set $A$ it is true that $A\subset \overline{A}$. 

So you have $\{x: f(x)&gt;0 \} \subset \overline{A}$. now, what about the points $x$ where there is equality, $f(x)= 0$ ? 

Assume $A\ne \emptyset$ ,try to show that $\overline{A} = \{x: f(x)\ge 0\}$ you can do that using the complement of the set, showing $\{x : f(x)&lt;0\}$ is open using the continuity of $f$.
for every set $A$ it is true that $A\subset \overline{A}$. 

So you have $\{x: f(x)&gt;0 \} \subset \overline{A}$. now, what about the points $x$ where there is equality, $f(x)= 0$ ? 

Assume $A\ne \emptyset$ ,try to show that $\overline{A} = \{x: f(x)\ge 0\}$ you can do that using the complement of the set, showing $\{x : f(x)&lt;0\}$ is open using the continuity of $f$.