Topology proof: dense sets and no trivial intersection

Question

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.

Proof:
[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.

could you give us your definition of a dense set? My definition coincides with that proposition.. — Exodd, Feb 28 '15 at 11:49
@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 $A$. — Kolmin, Feb 28 '15 at 11:51
How does $G\cap A'\neq\emptyset$ lead to a contradiction? You don't seem to say. — Michael Cotton, Feb 28 '15 at 11:55
It holds: $x\in\overline{A}\iff U_x\in\mathcal{T}:U_x\cap A\neq\varnothing$ — freishahiri, Mar 04 '15 at 15:45

Exodd · Answer 1 · 2015-02-28T12:29:57.643

1

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.

edited Feb 28 '15 at 12:29

answered Feb 28 '15 at 11:59

Exodd

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

score 1 · Accepted Answer · answered Feb 28 '15 at 12:13

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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 $G\cap A'\neq 0$.

answered Feb 28 '15 at 12:13

Michael Cotton

793

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?) – Kolmin Feb 28 '15 at 12:26
Yeah. You could write it that way as well. – Michael Cotton Feb 28 '15 at 12:33

Topology proof: dense sets and no trivial intersection

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