Suppose that $f,g:\mathbb{R} \rightarrow \mathbb{R}$ and are both continuous. For all $x \in \mathbb{Q}$, $f(x)=g(x)$. Show $f(x)=g(x)$ everywhere by using the epsilon-delta definition of continuity.
I know we can start off be taking c as a point not in the rationals and writing down what it means for f and g to be cts at this point but so far I can't figure out how to get any further.
Any help would be appreciated.
Choose $x\in\mathbb{R}$. Since the rationals are dense in the reals we can choose a sequence $(x_n)$ in the rationals which converges to $x$. Using continuity we have that \begin{equation} f(x) = \lim_{n\rightarrow\infty}f(x_n) = \lim_{n\rightarrow\infty}g(x_n) = g(x). \end{equation}
Let $h(x) = f(x) - g(x)$. Take $c\notin \Bbb Q$ and let $\varepsilon > 0$. Since $h$ is continuous, there is a $\delta > 0$ such that $|h(x) - h(c)| < \varepsilon$ for all $x$ in the open interval $(c - \delta, c + \delta)$. This open interval contains a rational point $q$, so $|h(q) - h(c)| < \varepsilon$. Since $q$ is rational, $h(q) = 0$, so $|h(c)| < \varepsilon$. Since $\varepsilon$ was arbitrary, $h(c) = 0$, i.e., $f(c) = g(c)$.
Assume without loss of generality that $f(c)-g(c)>0$ for some irrational $c$. By the continuity of $h(x)=f(x)-g(x)$, find an interval such that $h(x)>0$. In this interval there is some rational.
Can you make this rigorous?
