Using epsilon - delta to prove discontinuity of f(x)= non-integer part of x Let $f : R \rightarrow R$ be the non-integer part of $x$. Use the $ε-δ$  definition of continuity to show that $f$ is not continuous.
I have an idea of what this function looks like but am struggling with proving its discontinuity. I've been attempting it at the point where $c=1$, trying to fix $\epsilon >0$.
Thanks!
 A: As $f(x)$ is continuous in $x_0 \in \mathscr D(f)$ if $\forall \epsilon > 0, ~ \exists \delta > 0 ~ : ~ |x-x_0|<\delta \Rightarrow |f(x) - f(x_0)| < \epsilon$, we may say that $f(x)$ is not continuous if $\exists \epsilon > 0, \forall \delta > 0 ~ \exists x \in \mathscr D(f) ~ : ~ |x-x_0| < \delta , |f(x) - f(x_0)| > \epsilon$. It means that we need to find such $\epsilon$ and $x$ for which the last statement is true. We will try to prove that for $x_0 = 1$.
Let $\epsilon = 0.5$. Let's take $x<1$ and $x \rightarrow 1-0$, then $|x-x_0| \rightarrow 0$, so we can always find such $x$ that $|x-x_0| < \delta$ but the closer $x$ to $1$ is, the higher the difference between $f(x)$ and $f(x_0)$ is.
It's very important thing in such proofs that $x$ (I mean such $x$ for which the function is not continuous) depends on $\delta$, so as $|x-x_0| < \delta$ is the interval $x \in (x_0 - \delta, x_0 + \delta)$, we may take
$x = x_0 - \frac{\delta}{2}$, or $x = 1 - \frac{\delta}{2}$ if $\delta < 1$ and
$x = \frac 23$ if $\delta \geq 1$.
If $\delta < 1$ : as $f(x_0) = f(1) = 0$ and $f(1 - \frac{\delta}{2}) = 1 - \frac{\delta}{2} \Rightarrow |f(x) - f(x_0)| = 1 - \frac{\delta}{2}$. This is greater than $\epsilon = 0.5$ as $\delta < 1$.
If $\delta \geq 1$ : $|f(x_0) - f(x)| = |0 - \frac 23| = \frac 23$ and this is also greater than $\epsilon = 0.5$.
