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Let $ f: X rightarrow R^n$ continuous, where $ (X, d) $ is an arbitrary metric space. Prove or give a counterexample for the assertion that the set $M=xin X:f(x) ne 0 $ is an open set.

I'm thinking this might be fake. In the discret metric it seems true. So I'm thinking of an any metric $d$ and for the special case with $ X = mathbb R $ and $ n = 1 $.

I thought for the following functions:

1) $ f (x) = x Longrightarrow M = (- infty, 0) cup (0, infty) $. This set is opened.

2) $ f (x) = x ^ 2 Longrightarrow M $ is an open set.

So I conclude that if the roots of $ f $ is a finite quantity, we have that $ M $ is open. Now if it is an infinite contable quantity, for example for $ f (x) = sin x $, we have that $ M $ will be open.

So I thought of the following function
$$
f (x) = x sin left(dfrac 1 x right), x neq 0 ; mbox and ; ;
f(x)=0, x = 0.
$$

The roots of this function get closer and closer to zero because the roots $ x neq 0 $ are of the form $ x = dfrac 1 2k$ with $ k in mathbb Z $.
I thought I would have a counterexample here because I know I have points of $ M $ and points of $ mathbbR backslash M $ each time closer to each other. But I have not been able to conclude that this is indeed a counterexample.

$$M = f^-1 left( mathbbR^n setminus lbrace 0 rbrace right)$$

is open as the preimage of the open set $mathbbR^n setminus lbrace 0 rbrace$ by the continuous application $f$.

Let $f:XrightarrowmathbbR^n$ be continuous. Then $M$ is the complement of the set $M'=xin X:f(x)=0 $.

We need only prove that $M'$ is closed, since then by definition $M$ will be open.

Let $x_n_n=1^inftysubset M'$ be a sequence such that $x_nrightarrow x$ for $x in X$. To prove that $M'$ is closed, we prove that $x in M'$.

Indeed, we have that $f(x)=f(lim x_n)=lim f(x_n)=0$, since $f$ is continuous, thus $xin M'$ and $M'$ is closed.