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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 $\epsilon\geq 1$?

gbox
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  • Your question is a bit unclear. Do you mean why we made the choice $\varepsilon= \frac{1}{2}$? There is nothing special about $\frac{1}{2}$. Fixing any choice of $\varepsilon < 1$ works as it guarantees the tail sequence to have terms all equal. If $\varepsilon \ge 1$, then it may be the case that two terms are distinct. For example, take the discrete space of two points. ${-1,1}$, $(x_n)=((-1)^n)$. The proof does not work here if $\varepsilon$ was fixed to be some value $\ge 1$. – Bryan Shih Apr 13 '18 at 12:32
  • @CWL exactly, shouldn't we prove that for any $\epsilon$ the cauchy sequence converge? even for $\epsilon>1$ – gbox Apr 13 '18 at 12:33
  • This is true as you have shown all terms are equal: we have proved exists $N$, $x_n = X_N$ for all $n \ge N$. Let us pick arbitrary $\varepsilon > 0$, then $d(x_n,x_m) = d(x_N,x_N)=0 < \varepsilon$ for all $n,m \ge N$. – Bryan Shih Apr 13 '18 at 12:36
  • @CWL but the selection of $\epsilon$ is not arbitrary we only take $\epsilon<1$ that is what I do not get – gbox Apr 13 '18 at 12:39

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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.

egreg
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  • Yes, that is the issue, but still, maybe I have forgot some logical steps, by definition a sequence is Cauchy sequence if for all $\epsilon>0$... so should not we start with let there be $\epsilon$ without limitation (such as $\epsilon<1$) and prove that claim for the general $\epsilon$? – gbox Apr 13 '18 at 12:47
  • Ok I got it, we know that it is a Cauchy sequence, we just have to should that it converges, so we take $\epsilon<1$ because for this $\epsilon$ that Cauchy sequence converges – gbox Apr 13 '18 at 12:50
  • @gbox No, a Cauchy sequence either converges or doesn't. Choosing $1/2$ is a step towards showing the sequence is eventually constant. Remember: “a sequence converges if for every $\varepsilon>0$…”. A sequence doesn't “converge for every $\varepsilon>0$” (a sentence that makes no sense). – egreg Apr 13 '18 at 13:10
  • Ok, I got a step closer but still did not nailed it. If we take $\epsilon=2$ so the Cauchy sequence will not converge? – gbox Apr 13 '18 at 13:18
  • @gbox With $2$ in the first part you would conclude nothing, but the sequence would be convergent notwithstanding because it is a Cauchy sequence in a discrete space, as we proved by using $1/2$. – egreg Apr 13 '18 at 13:29