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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

R Morris
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    Your intuition is correct that this is a fine example for a non-complete metric space. What aspect of the proof do you have trouble with? – Roland Aug 19 '17 at 09:43
  • Im just unsure how far to go with it and how rigorous it should be @Roland – R Morris Aug 19 '17 at 09:52
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    If you are ok with the fact that $\frac 1 n$ converges to 0 in $\mathbb{R}$ for the distance $d$, then you do not need to add something else in my opinion. – nicomezi Aug 19 '17 at 09:55

1 Answers1

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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.

Sarvesh Ravichandran Iyer
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