At what points are $f(x)=[x]$ and $g(x)=x-[x]$ continuous? Given that $f(x)=[x]$ and $g(x)=x-[x]$. I want to determine the points where $f$ and $g$ are continuous.
Here is what I have done:
Let $f(x)=[x].$ It is obvious that if $b$ is an integer, then

$$\lim_{x\to b-} f(x)=\lim_{x\to b-}[x] =(b-1).$$
  Also, 
  $$\lim_{x\to b+} f(x)=\lim_{x\to b+}[x] =b,$$
  and 
  $$f(b)=[b] =b.$$

Since $\lim_{x\to b-} f(x)\neq \lim_{x\to b+} f(x), $ then $f$ is not continuous in $\Bbb{Z}$.
Let $g(x)=x-[x].$ It is obvious that if $b$ is an integer, then

$$\lim_{x\to b-} g(x)=\lim_{x\to b-}x -\lim_{x\to b-}[x] =b-(b-1)=1.$$
  Also, 
  $$\lim_{x\to b+} g(x)=\lim_{x\to b+}x -\lim_{x\to b+}[x] =b-b=0,$$
  and 
  $$g(b)=b -[b] =b-(b)=0.$$

Since $\lim_{x\to b-} g(x)\neq \lim_{x\to b+} g(x), $ then $g$ is not continuous in $\Bbb{Z}$.
How do I show that these functions are continuous for all $x\in\Bbb{R}\setminus \Bbb{Z}?$ 
 A: If $x \not \in \mathbb Z$ then there exists an $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon)$ is strictly between two integers (we can set $\epsilon$ to half the distance to the closest integer). Now $f$ is constant on $(x-\epsilon,x+\epsilon)$ and so is continuous at $x$. Try and use a similar argument for $g$.
A: As $f(x)+g(x)=x$, which is a continuous function, the floor and the fractional part share their discontinuities and there is no need to study both functions.
In a neighborhood that doesn't contain an integer, i.e. around any non-integer, the floor function is a constant, hence it is continuous.
Now at an integer, the floor values on the left and on the right differ by at least $1$, so there is a discontinuity.
A: $ \mathbb {R-Z}$ is the union of open intervals.
Thus for each point of $ \mathbb {R-Z}$ we have an open interval centered at the point and entirely included in     $ \mathbb {R-Z}$ .
Both functions are continuous on such an interval so they are continuous on the entire $ \mathbb {R-Z}$
