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Let $(E,d)$ be a metric space.

  • A sequence $(u_n)_n\in E^\mathbb{N}$ converges to $l\in E$ if for all $\varepsilon >0$, there exists $N\in \mathbb{N}$ such that $d(u_n,l)<\varepsilon$. I could also say that for any open subset $U$, $l\in U \Rightarrow \exists N, u_n\in U, \forall n\ge N$.

I understand that this notion of convergence is topological in the sense it depends only on the topology of $E$.

  • A sequence $(u_n)_n\in E^\mathbb{N}$ is said to be a Cauchy sequence if for all $\varepsilon >0$, there exists $N\in \mathbb{N}$ such that for all $p,q\ge N$, $d(u_p,u_q)<\varepsilon$. Here I could say for any open subset $U$, $0\in U \Rightarrow \exists N, u_n-u_p \in U, \forall n,p\ge N$.

Yet I do not understand why this notion of Cauchy sequence is not topological, that is we cannot define it onyl from the open sets of $E$ ?

I am trying to understand this with the fact that a convergent Cauchy sequence may not be converging in $E$. In that case, the limit could not be characterized with open sets of $E$. That would be the reason ?

José Carlos Santos
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Smilia
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1 Answers1

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On $(0,\infty)$, consider the usual distance $d_1$ ($d_1(x,y)=|x-y|$) and the distance $d_2$ defined by $d_2(x,y)=|\log(x)-\log(y)|$. These distances are equivalent (that is, they induce the same open sets), and therefore a sequence of elements of $(0,\infty)$ converges in $\bigl((0,\infty),d_1\bigr)$ if and only if it converges in $\bigl((0,\infty),d_2\bigr)$.

However, the sequence $\left(\frac1n\right)_{n\in\Bbb N}$ is a Cauchy sequence in $\bigl((0,\infty),d_1\bigr)$, but not in $\bigl((0,\infty),d_2\bigr)$.

So, the knowledge of the open subsets of a metric space $(X,d)$ is not enough to know its Cauchy sequences.

José Carlos Santos
  • 427,504