I have to decide if the function $\begin{Bmatrix} x-\frac{1}{2}\end{Bmatrix}$ is uniformly continuous in $(-\frac {1}{2},+\frac {1}{2})$ and in $[-\frac {1}{2},+\frac {1}{2}]$.
Analyzing the first case of $(-\frac {1}{2},+\frac {1}{2})$, we can say that the function is defined and continuous in $[-\frac {1}{2},+\frac {1}{2})$ and it presents a discontinuity in $x=\frac{1}{2}$ that can be eliminated putting $f(x=\frac{1}{2})= 1$.
With this position we can say that the function is defined and continuous in the closed and limited interval $[-\frac {1}{2},+\frac {1}{2}]$ and then for Weiestrass that the function is uniformly continuous in $[-\frac {1}{2},+\frac {1}{2}]$. Applying the estension theorem we can say that the function is uniformly continuous in $(-\frac {1}{2},+\frac {1}{2})$.
In the second case of the interval $[-\frac {1}{2},+\frac {1}{2}]$, it is not possible to say that the function is continuous because there is the discontinuity in $x= \frac{1}{2}$: for this reasons it is not a uniformly continuous function. If it is right (i'm not certain that my reasoning is 100% correct), it is strange how in the first case it was lecit to go to the conclusion passing through the fact the function was uniformly continuous in $[-\frac {1}{2},+\frac {1}{2}]$ and in the second one it was impossible.
In both cases, you are extending a continuous $f$ defined on some open interval $(a, b)$ to a continuous function $\tilde{f}$ defined on $[a, b]$. This is possible if the limits $\lim_{x\to b^-}f(x)$ and $\lim_{x\to a^+}f(x)$ exist. Once you have a continuous function on a compact set, in this case $[a, b]$, you can show that it is uniformly continuous. The restriction of a uniformly continuous function to subsets is also going to be uniformly continuous.
However, the function $\{x-1/2\}$ is not continuous on $[-1/2, 1/2]$, therefore not uniformly continuous.