# Investigating continuity of a function defined differently on the rationals and irrationals

For a function defined as follows:

$f(x) = \begin{cases} 1 & \text{if$x$is rational} \\ e^x & \text{if$x$is irrational} \\ \end{cases}$

How does one figure out the points at which f is continuous? Is there some kind of heuristic to figuring this out intuitively?

To be honest I was a little taken aback by the definition of the function. Naively I thought the function is likely continuous at $0$ since that's when the $e^x$ and $1$ become equal. Are there any other points?

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Heuristic: Draw a picture of $x\mapsto 1$, and $x \mapsto e^x$ and see where they cross. This is the only place they have a chance of being continuous. In fact, for any applicable problem, my first advice would be to draw a picture (but then again, I am an engineer). –  copper.hat Nov 8 '12 at 17:30
Do you know what "continuous" means? –  Chris Eagle Nov 8 '12 at 17:30

Let $a$ be any real number. There is always a sequence $\langle q_n:n\in\Bbb N\rangle$ of rational numbers converging to $a$, and along this sequence you have $\langle f(q_n):n\in\Bbb N\rangle\to 1$, since all of the $f(q_n)$ are equal to $1$. There is also always a sequence $\langle x_n:n\in\Bbb N\rangle$ of irrational numbers converging to $a$, and along this sequence you have $\langle f(x_n):n\in\Bbb N\rangle=\langle e^{x_n}:n\in\Bbb N\rangle\to e^a$. In order for $f$ to be continuous at $a$, these two limits must be the same. What does that tell you about $a$?
The two functions $f_1(x)= 1$ and $f_2(x)=e^x$ for themselves would be continuous. Then the function obtained by "mixing" them like in this problem is continuous exactly at those points $x$ where $f_1(x)=f_2(x)$ holds.
To see this, consider $x_0$ with $f_1(x_0)\ne f_2(x_0)$. Let $\epsilon = \frac12 |f_1(x_0)-f_2(x_0)|$. If $x_0$ is rational, you find irrational points $x$ arbitrarily close to $x_0$ such that $|f_2(x)-f_2(x_0)|<\epsilon$, hence $|f(x)-f(x_0)|=|f_2(x)-f_1(x_0)|\ge |f_2(x_0)-f_1(x_0)|-|f_2(x)-f_2(x_0)|=\epsilon$, i.e. $f$ is not continuous at $x_0$. The sam works with rational and irrational interchanged if $x_0$ is irrational.
On the other hand, if $f_1(x_0)=f_2(x_0)$, then for any $\epsilon>0$ there exists $\delta_1>0$ with $|f_1(x)-f_1(x_0)|<\epsilon$ if $|x-x_0|<\delta_1$ and $\delta_2>0$ with $|f_2(x)-f_2(x_0)|<\epsilon$ if $|x-x_0|<\delta_2$. Then with $\delta:=\min\{\delta_1,\delta_2\}$ we have $|f(x)-f(x_0)|<\epsilon$ if $|x-x_0|<\delta$.