# Set contains limit points of dense subsets

For a set $$A$$, denote $$A'$$ as the set of limit points of $$A$$. We define the closure of $$A$$ as the union of $$A$$ and the set of limit points of A. We write this as $$\overline{A} = A \cup A'$$.

WolframAlpha claims the following (http://mathworld.wolfram.com/Dense.html): In general, a subset A of X is dense if its set closure cl(A)=X.

I understand that if $$A$$ is dense in $$X$$ but $$A \not \subset X$$ then it is not necessarily true that $$\overline{A} =X$$.

But here we have that $$A$$ is dense in $$X$$ and $$A \subset X$$. So why is this fact $$\overline{A}=X$$ true? I tried proving it but am stuck. From raw definitions, let A be a dense subset of X where it is assumed that X is some metric space.

First, A is a subset of X so we have A $$\subset$$ X. Then A is dense in X by definition means that each $$x \in X$$ satisfies $$x \in A$$ or $$x \in A'$$ or both $$\implies X \subset \overline{A}$$.

To prove $$\overline{A}=X$$ we try to show inclusion the other way: that is, we want to show $$\overline{A} \subset X$$. Well, clearly $$A \subset X$$ so we need only show $$A' \subset X$$. Why must this hold? That is, why must a set contain all limit points of its dense subsets? My work so far is to choose an arbitrary point $$p \in A'$$ which requires $$\forall r>0: B(p,r) \cap A \not= \{\emptyset\}$$. Since $$A \subset X$$ then we claim $$\forall r>0: B(p,r) \cap X \not = \{\emptyset\}$$ but this only shows that $$p \in X'$$. I have not been able to use the definition of limit points to imply $$p \in X$$ as well.

So how come WolframAlpha makes this assertion that $$\overline{A} =X$$? Isn't it possible for $$A$$ to have a limit point that is outside of the metric space $$X$$, or would this end up contradicting the statement $$A$$ is dense in $$X$$?

• I think your confusion arises from the very definition of a limit point. The very definition of a limit point includes the ambient/parent space $X$ which contains $A$. Namely, limit points of $A$ are defined relative to the parent space $X$. So, when we say that $x$ is a limit point of $A$, we really mean that $x \in X$ is a limit point of $A$ relative to the space $X$. – rolandcyp Apr 2 at 23:33
• Implicitly, when we look at a space $X$ and consider its subsets, we think of $X$ as the "entire universe". Any topological claim about $A$ will implicitly be understood with respect to the space $X$ as we are not given a "larger space" $Y \supseteq X$ to work with. Unless you are given this larger space $Y$, it doesn't quite make sense to discuss points outside of the ambient space $X$. – rolandcyp Apr 2 at 23:35
• Thanks, I think I understand now. So just saying $A$ is dense would not make sense, right? Similarly, saying $A$ is dense in $B$ where $B$ is just another set is not technically correct either? We would need to specify that $A$ is dense in $X$ where $X$ must be a metric space. – Cethad Omes Apr 2 at 23:41
• Exactly, density and limit points are statements about the metric (or more generally, the topology). Then, it makes sense to say that $A \subseteq X$ is dense in $X$ (where $X$ is a metric space) if and only if $\operatorname{cl}(A) = X$. – rolandcyp Apr 2 at 23:42
• I see. But I've also seen mentions where the set $A$ is dense in some $X$ but $A$ is not a subset of $X$. How is this possible, if $X$ is taken to be an overarching metric space that serves as the entire universe? – Cethad Omes Apr 2 at 23:44