Non-constant continuous function having uncountably many zeros? I want to have an example of a non-constant function which has uncountably many zeros?
Is the following function continuous?
$f(x)=\prod\limits_{\alpha\in \mathbb{R\setminus Q}}(x-\alpha)$ If it so, it will be the required example.
 A: Your product does not converge for $x \in \mathbb{Q}$, so it does not define a continuous function. In fact the irrationals are dense, so a continuous function which is constant on the irrationals is constant everywhere.
If you are OK with a function which is locally constant in some regions, you can just take a continuous function with compact support, such as $f(x)=\max \{ 1-|x|,0 \}$. In some sense this feels like cheating to me, because it only achieves the goal by gluing a constant function to a non-constant function.
If you want a function which is nowhere locally constant and yet has uncountably many zeros, I can give a somewhat nonconstructive example: the trajectories of a Brownian motion have this property with probability $1$.
A: Let $C$ be any uncountable closed set, for example the Cantor set, and for every $x \in \mathbb{R}$, define $f(x) = d(x,C) = \inf\{d(x,c) : c \in C\}$. Since $C$ is closed, $f(x) = 0$ if and only if $x \in C$. Since $C$ is uncountable, we only need to verify that $f$ is continuous. To see this, fix $x,y \in \mathbb{R}$, and $c \in C$. Then by the triangle inequality, we have
$$d(x,c) \leq d(x,y) + d(y,c)$$
Taking the infimum over $c \in C$ gives us
$$f(x) \leq d(x,y) + f(y)$$
Similarly (interchanging the roles of $x$ and $y$ above), we have
$$f(y) \leq d(x,y) + f(x)$$
Combining these two inequalities, we conclude that
$$|f(x) - f(y)| \leq d(x,y)$$
which shows that $f$ is continuous (in fact, uniformly continuous).
P.S. I used the notation $d(x,y)$ above for the distance between $x$ and $y$, because the same proof works in any metric space that has an uncountable closed set $C$. In the case of $\mathbb{R}$, of course, we set $d(x,y) = |x-y|$.
