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Let $E$ be called dense in $\mathbb{R}$ if and only if $\text{int}(\mathbb{R} \setminus E)=\emptyset.$

Let $x \in\text{int}(\mathbb{R} \setminus E)=\emptyset$. Then for $\epsilon >0$, $(x-\epsilon, x+\epsilon)$ and $x \not\in E$. Hence $E$ is dense.

I'm not sure if this is the correct proof for the $\Leftarrow$ direction of the proof. Furthermore, I'm not sure how to proceed for the other direction of the proof.

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Why do you want to prove definition? –  mpiktas Sep 17 '12 at 13:43
    
So what exactly are the two definitions you want to show to be equivalent? –  Michael Greinecker Sep 17 '12 at 13:43
    
If $int(\mathbb{R}\backslash \rm E)=\emptyset$ is an alternative definition then what is the classical definition to you ? –  vanna Sep 17 '12 at 13:50
    
A set $E$ is dense in $X$ if every point of $X$ is a limit point, or a point of $E$. –  emka Sep 17 '12 at 13:51
    
@EMKA Now you just have to translate this in the language of "closure of sets", and you will see that Arthur's answer below answers your question. –  M Turgeon Sep 17 '12 at 13:55

2 Answers 2

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$\mathrm{Int}(\mathbb{R} \setminus A) = \emptyset$ means that $\mathbb{R} \setminus A$ has no interior points. In other words, every neighborhood of every point in $\mathbb{R} \setminus A$ contains a point in $A$. It follows that every point in $\mathbb{R} \setminus A$ is a limit point of $A$. Hence, $A$ is dense.

For the other way around, we know $A$ is dense. Every neighborhood of every point in $\mathbb{R} \setminus A$ must contain a point of $A$. Therefore, no point in $\mathbb{R} \setminus A$ can be an interior point of $\mathbb{R} \setminus A$. Hence $\mathrm{Int}(\mathbb{R} \setminus A) = \emptyset$.

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If we take it in the other direction. If we know that E is dense in $\mathbb{R}$. then every point in X is either a limit point or a point in E. So if we take $\mathbb{R}\setminus E$ and $\epsilon > 0$, then $(x-\epsilon, x+\epsilon)$ will continue a point $y$ such that $y$ is a limit point of point of $E$. So every open set $(x-\epsilon, x+\epsilon) \subset E$. Is this why the interior is empty? –  emka Sep 17 '12 at 14:18
    
@EMKA Your argument makes sense up to the "So every open set" part. $(x - \epsilon, x + \epsilon)$ has a point that belongs to $E$, but it's not fully contained in $E$. For $x$ to be an interior point of $\mathbb{R} \setminus E$, it needs to have an neighborhood fully contained in $\mathbb{R} \setminus E$, and this is not true as you can see in my edited answer. Does it make sense now? –  Ayman Hourieh Sep 17 '12 at 14:25

Your definition of a dense subset of a topological space is equivalent to the usual definition (i.e., closure is equal to the entire space).

It is a general fact for any topological space $X$ that $$X \setminus \mathrm{Int} ( A ) = \overline{ X \setminus A }.$$ Switching $A$ and $X \setminus A$ in the above we have that $X \setminus \mathrm{Int} ( X \setminus A ) = \overline{A}$, or $\mathrm{Int} ( X \setminus A ) = X \setminus \overline{A}$.

Therefore $E \subseteq \mathbb{R}$ is dense (in the usual sense) iff $\overline{E} = \mathbb{R}$ iff $\mathrm{Int} ( \mathbb{R} \setminus E ) = \mathbb{R} \setminus \overline{E} = \mathbb{R} \setminus \mathbb{R} = \emptyset$. (And $\mathbb{R}$ can be replaced everywhere by any topological space.)

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