How can a function be continuous if there are "gaps"? If you can, can you give the answer in as simple terms as possible?
Thank You
Asked
Active
Viewed 121 times
1
-
You can have continuity at a point without it being continuous everywhere. – JessicaK Apr 03 '16 at 01:55
-
I agree but from this function, is it only continuous between each integer? If so, what would that set look like? – user328121 Apr 03 '16 at 02:14
1 Answers
0
A function with "gaps," as you say, is not continuous. That said, a function can be continuous at some points, but not be continuous on the whole. The function in your link is continuous at some points, and discontinuous at others. The problem asks about points where it is discontinuous. Can you determine which points will be the locations of gaps, and at which the function will be continuous?
Plutoro
- 22,730
- 3
- 41
- 71
-
That's where my problem lies. If x is an integer, then f(x) will be x. If x is not an integer, f(x) will be 0. This means the if one was to draw the function, it would not be continuous since we will always be jumping between 0 and the integers. – user328121 Apr 03 '16 at 02:02
-
That's right, but if we just look between the integers, it is continuous at those points. It is discontinuous at each integer (except zero). – Plutoro Apr 03 '16 at 02:06
-
This may sound silly, but how does it prove it is continuous for when x is not an integer, since there are gaps in between each integer? – user328121 Apr 03 '16 at 02:11
-
-
-
At each nonzero integer point $x$, $f(x)=x\neq 0$. But there are points $y$ close to $x$ so that $f(y)=0$. – Plutoro Apr 03 '16 at 04:04
-
Normally in an introductory calculus class, you define a point $x_{0}$ to be continuous if $\lim_{x\rightarrow x_{0}^{+}} f(x) = \lim_{x\rightarrow x_{0}^{-}}f(x) = f(x_{0})$. Calculate those three quantities at an integer other than $x_{0}=0$ and see what you get. – JessicaK Apr 03 '16 at 06:23
-