[Math] Discrete Space is Complete Metric Space-About The Proof

metric-spaces

To prove that the discrete space is complete metric space we have to show that every cauchy sequence converges.
Why in proofs we have to use $\epsilon<1,\frac{1}{2}$ should not every cauchy sequence converge? even for $\epsilon\geq 1$?

Best Answer

You're probably confused about what you have to prove.

The task is proving that every Cauchy sequence converges. So, let $(x_n)$ be a Cauchy sequence. By definition of Cauchy sequence, there exists $N$ such that, for $m,n\ge N$, $d(x_m,x_n)<1/2$.

Since the metric is discrete, this implies that

for every $m,n\ge N$, $x_m=x_n$.

Therefore the sequence is eventually constant, hence convergent. Indeed, given $\varepsilon>0$, we have that, for every $n\ge N$, $d(x_n,x_N)=0<\varepsilon$.

Why $1/2$? Because it's good for the proof. Any positive number less than $1$ would have done as well.

Your confusion possibly comes from the dual usage of $\varepsilon$. Just not mentioning it in the first part should be sufficient for clearing up the matter.

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