0
$\begingroup$

In Rosenlicht's Introduction to Analysis, the author says, "...the knowledge of all the closed sets of a metric space is equivalent to the knowledge of all the open subsets. It is also true that the knowledge of all the open subsets of a metric space determines which sequences of points are convergent, and to which limits. For it is immediate from the definitions that the sequence $p_1, p_2, p_3,...$ converges to the point $p$ if and only if, for an open set $U$ that contains the point $p$, there exists a positive integer $N$ such that for any integer $n>N$ we have $p_n \in U$."

I have a question about each of these assertions. Firstly, why does the knowledge of all closed subsets give us all open subsets of a metric space? We can take the complement of each closed set to get an open set, but how do we know that each open set of the metric space is the complement of a closed set? A closed set is defined to be a set whose complement is open, but an open set isn't necessarily defined as the complement of a closed set, is it?

Secondly, why does the knowledge of all open sets give us all convergent sequences? I tried to prove the double implication. It's easy enough to prove "if $U$ is an open set, $p \in U$, and $\lim_{n\to\infty} p_n = p$, then there exists a positive integer $N$ such that for any integer $n>N$ we have $p_n \in U$" by stating that $U$ contains the open ball $\{x \in U: d(p,x) < \epsilon\}$, and when $n>N$, $p_n$ is in this ball and therefore $U$. But the reverse implication, "if $U$ is an open set, $p \in U$, and there exists a positive integer $N$ such that for any integer $n>N$ we have $p_n \in U$, then $\lim_{n\to\infty} p_n = p$" seems to be much harder. You can let $U$ contain the same open ball in an attempt to get $d(p, p_n) < \epsilon$, but you can only prove $p_n$ is in $U$, not necessarily in the open ball.

Thanks for reading.

$\endgroup$
  • $\begingroup$ The complement of the complement of any set is the set because if an element is not among the things that are not in the set it must be in the set. $\endgroup$ – John Douma Nov 5 '15 at 19:04
  • $\begingroup$ A set is open iff its complement is closed. So each open set $U = (U^c)^c$ is the complement of a closed set $\endgroup$ – zhw. Nov 5 '15 at 19:09
1
$\begingroup$

First, it is true that any open set is the complement of a closed set: let $U$ be open. Then, by definition, $F:= \complement U$ is closed. Since $U = \complement(\complement U) = \complement F$, it follows that $U$ is the complement of a closed set.

Second, the statement of the book is a little confusing as stated. As you saw, it's not only hard to prove, but in fact false that a sequence $p_n$ converges if for some open set $U$, $p_n \in U$ for all sufficiently large $n$. Indeed, if this were true, all sequences in any space $X$ would converge, since we could just take $U = X$!

The statement needed is: A sequence $\{p_n\}_n$ in a metric space $X$ converges iff for any open set $ U \subseteq X$, there is some $N$ (depending on $U$) such that for all $n > N$, $p_n \in U$.

To see this, just note that open balls are an example of open sets, and as you observed, convergence is the same thing as the sequence eventually staying within any sufficiently small open ball (essentially by definition).

Note: For more general spaces $X$ which may not have the structure of a metric space, we can still define the notion of an open set (in which case we call $X$ a topological space, of which metric spaces are among the nicest examples), with closed sets being exactly the complements of open sets as above. We may define convergence of a sequence by the above property, which doesn't require a metric to state. Some requirements on $X$ are required to make sure this definition has the nice properties you expect from the metric space case, but the definition itself works in complete generality.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That makes much more sense! Thank you for explaining. $\endgroup$ – Wyatt Gregory Nov 5 '15 at 19:05

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.