Prove that $f(x)=\lfloor x \rfloor $ is discontinuous

Question

I understand the first case. Since, $\lim_{x \to c}[x]=f(c)=[c]$ $f$ is continuous for all $x \in \Bbb{R-Z}$

My problem in understanding is with Case II, here, $c \in \Bbb{Z}$.

We take a very small number $r>0$ such that $[c-r]=c-1$ whereas $[c+r]=c$. I don't understand this. I know this is related to the property of the greatest integer function but can anyone explain this in simple words and help me understand this? Perhaps give an example? (I'm a highschool student)

What is it you don't understand? $[c-r]=c-1$ seems clear enough. Take a concrete example if it's difficult: do you see how $[3-0.00001]=2$? — Arthur, Dec 27 '17 at 13:51

score 2 · Accepted Answer · answered Dec 27 '17 at 13:50

2

$[0-0.0001]=-1$ while $[0+0.0001]=0$; $[0-0.0000001]=-1$ while $[0+0.0000001]=0$. A change in the argument by a tiny quantity results in a finite change of the function. And reducing the $\epsilon$ doesn't help.

A plot makes it obvious.

answered Dec 27 '17 at 13:50

Thank you; your answer made me realise how trivial my question was! – Raknos13 Dec 27 '17 at 14:05

score 1 · Answer 2 · answered Dec 27 '17 at 13:51

1

Example:

If $c = 5$, we can take $r = 0.004$, such that: $$\lfloor c-r \rfloor = 4 \, \text{ and } \lfloor c+r \rfloor = 5$$

answered Dec 27 '17 at 13:51

Prove that $f(x)=\lfloor x \rfloor $ is discontinuous

2 Answers2