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 $X$?

  • $\begingroup$ 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 $X$. $\endgroup$ – rolandcyp Apr 2 at 23:33
  • $\begingroup$ 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 $X$. $\endgroup$ – rolandcyp Apr 2 at 23:35
  • $\begingroup$ 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. $\endgroup$ – Cethad Omes Apr 2 at 23:41
  • $\begingroup$ 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 $\operatorname{cl}(A) = X$. $\endgroup$ – rolandcyp Apr 2 at 23:42
  • $\begingroup$ 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? $\endgroup$ – Cethad Omes Apr 2 at 23:44

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