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Question

Let's say we have a function: $$f(x)= \left\{ \begin{array}{c} 3x,\ if\ x\leq2 \\ floor(x)+4,\ if\ x>2 \end{array} \right. $$

How do we prove continuity over $(-\infty,2]$ and discontinuity over $Z \cap (2, \infty)$?

This is where we are so far:

To prove that f(x) is continuous over p = $(-\infty,2]$, we must prove that $\forall$ $\epsilon$ $\in$ $(0,1)$, $\exists$ $\delta > 0$ s.t. $\forall$ q $\in$ $\mathbb{R}$, $|p - q|$ < $\delta$ $\implies$ $|f(p) - f(q)|$ < $\epsilon$.

For continuity over $(-\infty,2]$, there are two cases to handle:

Case 1: $q \leq 2$

$|f(p) - f(q)|$ < $\epsilon$

= $|3p - 3q|$ < $\epsilon$

= $|p - q|$ < $\frac{\epsilon}{3}$

Choose $\delta = \frac{\epsilon}{3}$. Case proven.

Case 2: $q > 2$

$|f(p) - f(q)|$ < $\epsilon$

= $|3p - (floor(q) + 4)|$ < $\epsilon$

= $|3p - floor(q) - 4|$ < $\epsilon$

Where do I go from here?

For continuity over $(2, \infty]$, there are likewise two cases to handle:

Case 1: $q \leq 2$

$|f(p) - f(q)|$ < $\epsilon$

= $|floor(p) + 4 - 3q|$ < $\epsilon$

Where do I go from here?

Case 2: $q > 2$

= $|floor(p) + 4 - (floor(p) + 4)|$ < $\epsilon$

Where do I go from here?

Thank you!

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    There are only discrete points of discontinuity, one at each integer, for the "floor" function over $(2, \infty)$. The function is not discontinuous everywhere, only at these discrete points. It is continuous in between each pair of integers. Also, I am rather well known for liking to discuss things at length and working through every tiny detail. And even I think that you are overcomplicating things on your $\epsilon -\delta$ proof. Look at some models in good texts and cut way way back. – 2017-02-14
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    Thank you! I'll fix the problem to reflect that it is only discontinuous over Z greater than 2. As for detail - our professor wants us to go into this level of detail. How would you go about proving that the function is discontinuous at each integer? – 2017-02-14
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    I have to get off the computer now. Not ignoring you but real life. Somebody on here should be able to give you some more tips. – 2017-02-14
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    No, it is *only* discontinuous at *integers* greater than 2. Be exact. Say what you mean and mean what you say. (Bonus points for source of that quote). – 2017-02-14

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