Closure of a set defined using a function

Question

I have the following set in $\mathbb{R}^2$:

$$ S=\{(x_1,x_2)\in\mathbb{R}^2 : f(x_2)<0\} \neq \varnothing $$ for some continuous function $f\colon\mathbb{R}\to\mathbb{R}$. Can we say that the closure of $S$ is given by $$ \overline S=\{(x_1,x_2)\in\mathbb{R}^2 : f(x_2)\leq 0\} $$ Please help me as I am new to set theory...

Answer 1

(This answer addresses the original question.)

As @TPace suggests, if $f$ admits a point of discontinuity, then it's not necessarily true.

Consider the function $f(x) = \begin{cases} -1 & x < 0 \\ 1 & x \ge 0\end{cases}$. Since $0$ is not in the range of $f$, $f(x) < 0$ is equivalent to $f(x) \le 0$. Can you continue?

\begin{align} S &= \{(x_1,x_2) \mid f(x_2)<0\} \\ &= \{(x_1,x_2) \mid f(x_2) \le 0\} \\ &= \Bbb{R} \times (-\infty,0) \\ \bar{S} &= \Bbb{R} \times (-\infty,0] \ne S \end{align}

Answer 2

Let f be the continuous zero function. Thus S is empty.
However P = { (x,y) : f(y) <= 0 } is the whole plain.
Consequently P /= $\overline S.$

A continuous counter example with a not empty S is
f(x) = 0 if x < 0; f(x) = -x if 0 <= x.

Are you afraid of asking a question directly?
For example, "Is the closure of S ...?"