3
$\begingroup$

I am very new to analysis. To show a space $X$ is complete, they always begin with, such as, "Let $(x_n)$ be a Cauchy sequence in $X$". But I am confused that why we can must find a Cauchy sequence in $X$, without knowing any property of $X$. In other words, there must exist Cauchy sequence in any space $X$, why?

I know Cauchy sequences are those whose elements become closer and closer in terms of a norm, as the label goes larger and larger. While I think nothing guarantees such a sequence exists in arbitrary space.

For example from a textbook: (Please pay attention to my remarks)

To show a linear subspace $M$ of a Banach space $(X, \|\cdot\|)$ is complete if and only if it is closed.

Complete $\Rightarrow$ Closed:

  1. Let $x \in \overline{M}$, then there is a sequence $(x_n)$ in $M$ such that $\|x_n - x\| \to 0$ as $n \to \infty$. (Remark: Why is there such a sequence? Why does it guarantee that we can find such a sequence?)
  2. Since $(x_n)$ converges, it is Cauchy. (Remark: This is true if $M$ is a normed linear space. A subspace of a normed space must be normed with the same norm?)
  3. Completeness of $M$ guarantees the existence of an element $y \in M$ such that $\|x_n - y\| \to 0$ as $n \to \infty$.
  4. By uniqueness of limits, $x = y$.
  5. Hence $x \in M$ and consequently $M$ is closed. (Remark: Which one is the definition of closed set: Its complement set is open, or, the limit points of all its sequences are contained in it? If the later, $(x_n)$ is "any" sequence rather than "a" sequence in step 1, and consequently $x$ here is every limit point?)

Complete $\Leftarrow$ Closed:

  1. Let $(x_n)$ be a Cauchy sequence in $M$. (Remark: Why does such a Cauchy sequence exist and can be found? I mean, who can guarantee this? Indeed, here it means "a" Cauchy sequence or "any" Cauchy sequence?)
  2. Then $(x_n)$ is a Cauchy sequence in $X$.
  3. Since $X$ is Banach and thus complete, there is an element $x \in X$ such that $\|x_n - x\| \to 0$ as $n \to \infty$.
  4. But then $x \in M$ since $M$ is closed. (Remark: Is this by definition of closed set I mentioned?)
  5. Hence $M$ is complete. (Remark: It seems not "a" but "every" Cauchy sequence in step 1?)

As I said, I am very new to analysis. I am really hoping someone can help me to review and answer my remarks carefully. Thank you in advance for your patience.

$\endgroup$
  • $\begingroup$ How would you prove a statement of the form "$\forall z\in A: x\in B$", where $A,B$ are sets? The usual way is to start with "Let $z\in A$ be given." This also works if $A$ is empty, and it means that $z$ is arbitrary, not just specific. $\endgroup$ – supinf Aug 17 '17 at 10:42
4
$\begingroup$

In any non-empty space $X$ there are Cauchy sequences. Just take some $x\in X$ and define $(\forall n\in\mathbb{N}):x_n=x$.

Concerning the question about why $x\in\overline M$ implies thatthere is a sequence $(x_n)_{n\in\mathbb N}$ that converges to $x$, that's because that's one of the properties of the closure (some authors take it as the definition of closure).

And, yes, when we talk about a subspace of a normed space, the default assumprion is that the norm is the same as that of the original space.

Finally, concerning the definition of closed set, it's up to you to tell us which definition that textbook of yours contains. It's not as if all textbooks use the same definition.

$\endgroup$
3
$\begingroup$

If there are no Cauchy sequences, the space is automatically complete, so there is nothing to prove.

$\endgroup$
1
$\begingroup$

You must show that:

$(x_n)$ cauchy sequence in $X \Rightarrow (x_n)$ converges

The implication $p \Rightarrow q$ is true if $p$ is false, so if there are no cauchy sequences, the statement is trivially true.

Of course, in a non empty space $X$, any constant sequence is a cauchy sequence, so you can always find a cauchy sequence.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.