Neighbourhood property of continuity Let $c\in\Bbb R$ and a function $f:\Bbb R\to\Bbb R$ is continuous at c.
If for every positive $\delta$ there is a point $y\in(c-\delta, c+\delta)$ such that $f(y)=0$, prove that $f(c)=0$.
 A: Hint
If $f(c) \neq0$, you would be able to build a sequence $\{x_n\}$ with $x_n \neq c$ converging to $c$ such that $f(x_n) = 0$ for all $n$. A contradiction if $f$ is supposed to be continuous.
A: Apply the property to $$\delta = \frac{1}{n}$$
There exists a point $y_n \in \left(c- \frac{1}{n}, c + \frac{1}{n}\right)$ such that $f(y_n)=0$.
But because $$c- \frac{1}{n} < y_n < c + \frac{1}{n}$$
by the squeeze theoreme, $(y_n)$ tends to $c$ when $n$ tends to $+\infty$. Letting $n$ tend to $+\infty$ in the equality $f(y_n)=0$ gives you, by continuity of $f$, that
$$f(c)=0$$
A: $f$ is continuous at $c$, so $f(c)$ exists. If $f(c)$ is not $0$ then  then the limit of $f(x)$ as $x \to c$  does not exist. This is because $every$ neighbourhood around $c$ contains a point $y$ such that $f(y) = 0$, and so $\nexists \delta>0$ such that $x \in (c-\delta, c+\delta) \implies f(x) \in (\frac{1}{2}f(c),\frac{3}{2}f(c)) \quad $ (i.e. $f$ is not continuous at $c$, a contradiction).
But since $f$ is continuous at $c$, $f(c)$ does exist and therefore must be $0$.
