I was wondering if this proof of this basic topological result concerning the closure works.

Proposition: Let $A \subseteq (X,\tau)$. Then, $A$ is dense in $X$ if and only if every non-empty open subset of $X$ intersects $A$ non trivially.

[only if] Assume that $A$ is dense in $X$. This means that $A \cup A' = X$, where $A'$ denotes the set of all the limit points of $A$. Let $G \in \tau$ be non-empty. Thus, $G \subseteq (A \cup A') = X$. Hence, either $G \cap A \neq \varnothing$, which establishes the result, or $G \cap A' \neq \varnothing$, which leads to a contradiction.

[if] Assume that $A$ is not dense in $X$. Thus, there is a $x^* \in X \setminus (A \cup A')$. Notice, that $X \setminus (A \cup A')$ is open, because $A \cup A'$ is closed, and it is not empty. Hence, we can conclude by realizing that $X \setminus (A \cup A') \cap A = \varnothing$, which completes the proof. $\square$

It is one of the first proofs I attempt in topology, and I am not planning to see how it is proved in the book I am reading (unless you answer me that my proof is completely wrong).

Thus, I am really looking forward to any feedback, also contentwise.

Thank you for you time.

  • $\begingroup$ could you give us your definition of a dense set? My definition coincides with that proposition.. $\endgroup$ – Exodd Feb 28 '15 at 11:49
  • $\begingroup$ @Exodd: A set $A$ is dense in $X$ if $\bar{A} = X$, where $\bar{A} = A \cup A'$, and $A'$ denotes the set of all limit points of $A$. $\endgroup$ – Kolmin Feb 28 '15 at 11:51
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    $\begingroup$ How does $G\cap A'\neq\emptyset$ lead to a contradiction? You don't seem to say. $\endgroup$ – Michael Cotton Feb 28 '15 at 11:55
  • $\begingroup$ It holds: $x\in\overline{A}\iff U_x\in\mathcal{T}:U_x\cap A\neq\varnothing$ $\endgroup$ – C-Star-W-Star Mar 4 '15 at 15:45

I think what you want to say instead is that either $G\cap A\neq 0$ which establishes the result, or else we would have both $G\cap A'\neq 0$ and $G\subseteq X\setminus A$. But since G is open, $G\subseteq X\setminus A\Rightarrow G\subseteq X\setminus \bar{A}$ which contradicts $G\cap A'\neq 0$.

  • $\begingroup$ Thanks a lot for the point. I was just very sloppy. Thinking more carefully I realised (correct me if I am wrong), that I can establish the result also by writing that, if I assume that $G \cap A = \varnothing$ and $G \subseteq A'$, I get a contradiction by the definition of limit point, because for every open set $U \in \tau$, a point $x$ is a limit point of $A$, iff there is another point $y \neq x$ s.t. $y \in U \cap A$. Thus, it cannot be that $G \cap A = \varnothing$. (Right?) $\endgroup$ – Kolmin Feb 28 '15 at 12:26
  • $\begingroup$ Yeah. You could write it that way as well. $\endgroup$ – Michael Cotton Feb 28 '15 at 12:33

Why does $G\cap A'\ne \emptyset$ leads to a contradiction? We know that $$G\cap A = \emptyset \implies G\subseteq A' - A$$ but you have to say something more in order to reach the absurd (unless you're using some propositions you proved earlier)

The rest is Ok.

  • $\begingroup$ The limit points of $A$ can overlap with $A$, so even that doesn't seem quite right. $\endgroup$ – Michael Cotton Feb 28 '15 at 12:05
  • $\begingroup$ It wasn't wrong that $G\subseteq A'$, but I suppose that we need more to reach the absurd, thanks $\endgroup$ – Exodd Feb 28 '15 at 12:09
  • $\begingroup$ Ah. With the minus there, it's clearer what you meant. $\endgroup$ – Michael Cotton Feb 28 '15 at 12:21

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