Not complete metric spaces I am trying to show that the metric space:
$(X,d)$, where $X=(0,1] 
$ and $d(x,y)=|x-y| ∀x,y ∈ X$ is not complete.
My thoughts are that if I define a Cauchy sequence such as $(X_k)_{k>=1} = \frac 1k$ which tends to $0$ as k approaches infinity then this could be a proof by counter example as $0$ is not in X
any thoughts would be great 
 A: You are right in thought, it's nice if you want to be rigorous. Any person with even basic exposure to real analysis will understand your unrigorous proof  immediately.
As for rigor, there isn't much to say. I'll write it in steps :
1) To  show that $(0,1]$ is not complete under the absolute value metric, it is enough to find a Cauchy sequence in $(0,1]$ which does not converge in $(0,1]$.
2) Define $a_k = \frac 1k$ for all $k \in \mathbb N$. Clearly, $0 < a_k \leq 1$ for all $k$, hence $\{ a_k\} \subset (0,1]$.
3) $|a_k - a_l|= |\frac 1k - \frac 1l| = |\frac{l-k}{kl}|$. Given $\epsilon > 0$, choose $M$ so large that $\frac 1M < \frac \epsilon 2$. Then, 
$$
l,k > M \implies \left|\frac 1l - \frac 1k\right| \leq \left|\frac 1l\right| + \left|\frac 1k\right| < 2\frac \epsilon 2 < \epsilon
$$
So the  sequence $\{a_k\}$ is Cauchy.
4) Note that  $|a_k - 0| = |a_k| = \frac 1k$ can be made less than $\epsilon > 0$ by simply taking $k > \frac 1 \epsilon$. Hence, $a_k \to 0$. However, $0 \notin (0,1]$, hence $\{a_k\}$ is a Cauchy sequence which is not convergent in $(0,1]$.
A rigorous proof, if you like. You will not need this much detail  in future.
