Why are these two expressions equivalent?

Let $X$ be a topological space and let $A\subseteq X$. Can someone please tell me 1) what the difference between these two expressions is 2) if they are equivalent (if they aren't can someone give me two counterexamples in which in turn each expression holds and the other doesn't?) 3) how one would formulate the first expression in words ?

$\forall x \in X: \ \ \forall (x_n)_n$ in $A:\ x_n \rightarrow x \ \Rightarrow x \in A$

$\forall x \in X: \ \ \exists (x_n)_n$ in $A:\ x_n \rightarrow x \ \Rightarrow x \in A$

(The motivation behind this question is that in a course in topology which I took, the first expression was used in the following theorem that stated that $A$ was closed iff the first expresion would hold:

"Let $X$ be a first countable topological space, $A\subseteq X$ and $x \in X$. Then we have: $A$ is closed $\Leftrightarrow$ $\forall (x_n)_n$ in $A,\ x_n \rightarrow x \ \Rightarrow x \in A$ "

And the second expression was used in the following theorem that stated that what the closure of $A$, $c(A)$, would look like:

"Let $X$ be a first countable topological space, $A\subseteq X$ and $x \in X$. Then we have: $c(A)= { x\in X |\ \exists (x_n)_n \textrm{ in }A:\ x_n \rightarrow x }$"

So my guess was, that the above expressions should also be equivalent formally (if we take the st theoretic notation from the second theorem away), since they express the same thing - so I think the proof of the equivalence should not use the fact that our topological space is actually first countable and one of the expression is identical to the fact that $A$ is closed.

Please take note, that this is how the theorems were written down on the blackboard; that is all I have (even if it may be incorrect).

Since I'm not very good with formal notation I may have written the above expression falsely.)

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Note that if $A$ is not first countable this is not true, and you'd need nets and not sequences. – Asaf Karagila Sep 21 '11 at 8:30
The second sentence is odd and needs some parenthesis (around "$\exists (x_n)$ in $A$ : $x_n\to x$", I guess). As written, it is equivalent to: for all $x$ in $X\setminus A$, there exists $(x_n)$ in $A$ such that $x_n$ does not converge to $x$... – Did Sep 21 '11 at 8:42
I have deleted my answer in waiting for a respond on Didier's comment above before I put efforts in correcting my answer. – Asaf Karagila Sep 21 '11 at 8:48
@Asaf, yes. My now deleted comment on your now deleted answer referred to the sentence $\forall x\in X,\exists (x_n)$ in $A,[x_n\to x\implies x\in A]$. Note that in the absence of parenthesis, this is the usual way to read this, I believe. – Did Sep 21 '11 at 8:50
Because (1) in the absence of parenthesis the convention is to add them "from the end", that is, as in my comment to @Asaf above, (2) if you understand the sentence like that, it is equivalent to what I wrote in my comment to you, (3) what I wrote there is true for every nonempty $A$. – Did Sep 21 '11 at 9:53

The punctuation of the formulas in the question is very odd, and probably not enough to give a unique way of reading the formula. But here is what the formulas mean if they are read with the usual conventions for quantifiers.

The formula $$(\forall x \in X)(\forall (x_n) \in A)[ (x_n) \to x \Rightarrow x \in A]$$ says that $A$ contains all its sequential adherent points.

The formula $$(\forall x \in X)(\exists (x_n) \in A)[ (x_n) \to x \Rightarrow x \in A]$$ says that for all $x \in X$ there exists a sequence $(x_n)$ of points of $A$ such that if $(x_n) \to x$ then $x \in A$. With the $T_1$ axiom this is simply equivalent to $A$ being nonempty:

• If $A$ is nonempty, suppose I am given an $x \in X$. If $x \in A$ let $x_n = x$ for all $n$. Then $(x_n)\to x$ and $x \in A$, so the formula is true. Otherwise, $x \not \in A$. Fix some point $a \in A$ and let $x_n = a$ for all $n$. Then $(x_n) \to a$, so $(x_n) \not \to x$ by $T_1$, which means the formula is true. Either way, if $A$ is nonempty there is a sequence $(x_n)$ that works.

• If $A$ is empty, let $x$ be a point of the space. Then there is no sequence $(x_n)$ of points in $A$, so the formula is false.

It is true that the first formula, $$(\forall x \in X)(\forall (x_n) \in A)[ (x_n) \to x \Rightarrow x \in A]$$ is equivalent to the following formula $$(\forall x \in X)[((\exists (x_n) \in A)[ (x_n) \to x]) \Rightarrow x \in A]$$

This is an instance of a general scheme: $$((\exists z)\Psi) \Rightarrow \Phi) \Leftrightarrow (\forall x)(\Psi \to \Phi)$$ which holds if $z$ is not mentioned in $\Phi$ (in this case $z$ is $(x_n)$).

Perhaps the point I am making is that the parentheses are very important, because it makes a difference whether the quantifier is in the hypothesis of a conditional statement, or whether the entire conditional statement is in the scope of the quantifier.

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nitpick: $\forall (x_n) \in A$ makes no sense (a sequence is not an element of $A$), we do need a more cumbersome notation like $\forall (x_n)$ s.t. $\forall n: x_n \in A$, or $\forall (x_n)$ s.t. $\{x_n : n \in N \} \subset A$ etc. – Henno Brandsma Sep 21 '11 at 11:31
That would be better, I agree, but I was trying to follow the notation of the question. Writing $(x_n) \in A$ or $(x_n) \subseteq A$ on blackboards is not uncommon in my experience, although I would avoid it in any sort of written document. In any case, in my answer, please read $(\exists (x_n) \in A) \Phi$ as an abbreviation for $(\exists x)[ (x \colon \mathbb{N} \to A) \land \Phi]$, and similarly for $(\forall (x_n) \in A)$. – Carl Mummert Sep 21 '11 at 11:36
@Carl Elucidation at last...thanks Carl. You really have a talent for explaining! Even if wasn't able to formulate precisely in my question what I didn't understand, you deciphered it correctly and pinpointed my problem. – temo Sep 24 '11 at 18:09

As a caveat: the equivalence to closedness and closure that you mentioned holds in so-called first countable spaces, so maybe the course you took was on e.g. metric spaces (as is common; my first topology course was called "metrische topologie" (metric topology) as well).

The first expression says: $A$ is closed under sequential limits: for all sequences from $A$ (i.e. a sequence all of whose elements are from $A$) that converge, the limit is in $A$ as well. For e.g. metric spaces (see above) this is indeed equivalent to A being closed.

The second set $\lim(A) = \{ x \mid \exists (x_n) : \forall_n x_n \in A \wedge x_n \rightarrow x \}$ is the set of all limits of sequences from $A$. For metric spaces (etc. ) this indeed equals the closure of a set $A$.

The relation can be seen as follows: the closure of $A$, $\overline{A}$ must be a closed set, so one needs to show the following (which is indeed true): $\lim(A)$ must be closed under sequential limits itself, so if a sequence of points, all of which limits of sequences from $A$, converges to a point $x$, then $x$ too is a limit of a sequence from $A$.

This follows quite easily (try showing it yourself) in metric spaces e.g. It's not true in general spaces though.

If we have shown that fact, namely that $\lim(A)$ is closed (using limits), then it follows that it is the closure of $A$: it contains $A$, using constant sequences, and limits from sequences from $A$ are always in the closure of $A$, in any space, so it's the smallest closed superset of $A$.

But the relation is not purely logic, as you maybe think, it does involve some topology arguments, and does not hold in all topological spaces.

Also, the first is a property of a given set $A$ ($A$ is closed under limits), while the second (in the set form) defines another set defined from $A$. If this new set $\lim(A)$ equals $A$, then (in metric spaces etc.) the set is closed.

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In my reading, the second formula of the question does not say $(\exists (x_n))[(\forall n [x_n \in A_n]) \land (x_n) \to x]$. It says $(\exists (x_n))[(\forall n [x_n \in A_n]) \land ((x_n) \to x) \Rightarrow x \in A)]$. The first formula in this comment says that the sequence converges to $x$. The second merely says that if the sequence converges to $x$, then $x$ is in $A$. – Carl Mummert Sep 21 '11 at 11:08
@Carl: From what he was saying about the meaning (nl one means $A$ is closed, the other the closure of $A$) I inferred my interpretation of the interpunction to use, namely the one you mentioned first: which is a predicate on $x$, namely that there is a sequence from $A$ converging to $x$. He then forms the set of all such $x$ (comprehension). But admittedly he is somewhat unclear. – Henno Brandsma Sep 21 '11 at 11:26

If I'm not misunderstanding the correct meaning of your questions it seems to me quite obvious that they are not the same sequence in general but you have to clarify what you really mean with them.

In particular the second sentence, as I read it, means that every time you have a sequence of elements in $A$ converging at some point $x$ then this point is in $A$. This is true if you have a closed set but in general is not (take $1/n$ in $(0,1)$).

The first sentence (again as I read it) means that if every sequence in $A$ converges to the same point $x$ then this point is in $A$. This seems a little bit confusing because you can always find in a set which as more than one element two different sequence (the constant ones) converging to different points and so the hypotesis of your impications is almost always false.

---Edit--- From the new edit point of view is seems that the the sequences should be red as:

1. characterize closed set among others via convergent sequences: every time you've got a convergent sequence the limit is a point of the set in which the sequence lays.
2. characterize the closure of a set $A$. In particular you get the definition of closure (as you have said) taking the second $A$ in the sentence being $c(A)$.

Now, turning to the equivalence if you are asking about logic equivalence clearly there is a gap between the two sentences (not only referring to the $\forall$ and $\exists$ sa my previous answer was about): in particular you are comparing the second part of an equivalence (as in the characterization of a closed set) and a part of the definition of a set. So imo strictly logically speaking you cannot get an equivalence.

Topologically you have to care about separatedness axioms like said above and so on...

I hope I've undestood (better now) your question and that my answer is a little claryfing for you.

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