First, let's examine these values individually.

For $a$, using the fact that the base of the logarithm $\frac{1}{2}$ is less than 1, which makes the logarithm function decreasing, we compare $a$ with the logarithm of a nearby easy-to-calculate number, which is $4$:

$$a = \log_{\frac{1}{2}}5 2^{0} = 1.$$

Now, since $a 1$, we can conclude:

$$a < -2 < 0 < b < 1 < c,$$

therefore, $a < b < c$ holds.

Therefore, the correct answer is:

$\boxed{A:}$ a < b < c}.