Question regarding continuous functions.

Question

Question: Suppose $f:\mathbb{R}^n\rightarrow\mathbb{R}$ Show that the set $\{x\in\mathbb{R}^n \mid f(x)=0\}$ is a closed set.

My solution (most probably wrong): a function is continuous iff the inverse image of every closed set is closed. Thus we must show that for $x\subset M$ the inverse image of $X$ under $f^{-1}$ This however means $(f^{-1})^{-1}X=f(X)$ is closed.

So... I am unsure if the proof I am using is acceptable given the question. If there are any adjustments that I should make please let me know.

Answer 1

If you're not comfortable with point-set topology use the sequential argument: to show that $X = f^{-1}\{0\}$ is closed, we have to show that every limit point of $X$ is in $X$. Let $x$ be a limit point of $X$, then there is a sequence of points $x_n$ which converges to $x$. Now since $f$ is continuous, it preserves limits. More precisely, since $\lim_{n \to \infty }x_n =x$, we have that $f(x) = f(\lim_{n \to \infty}x_n) = \lim_{n \to \infty}f(x_n)$. Since $x_n \in X$, we have that $f(x_n) = 0$ for each $n$. Therefore $\lim_{n \to \infty}f(x_n) = 0$ and hence $f(x) = 0$. So $x \in X$. Therefore $X$ is closed.

Answer 2

As you said, the inverse image of a closed set under a continuous function is closed, your set is the inverse image of the singleton $\{0\}$, which is closed in $\mathbb{R}$ with the usual topology, hence its inverse image is closed in the euclidean space $\mathbb{R}^n$.