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I am working on a problem I found that asks whether the set $S = \{x \in \mathbb{R} \mid x \ge 0\}$ is dense in $\mathbb{R}$.

The theorem I have been using states the following:

"$S$ is dense in $\mathbb{R} \iff \forall a,b \in \mathbb{R}$ with $a < b$, then $\exists x \in S$ such that $x \in (a,b)$."

Now my logic from what I have so far is basically that for any $a$ and $b$ you give me, I can find the midpoint between $a$ and $b$, which will consistently give me a positive real number. This feels like I'm just constructing sentences and not really "proving" it, however.

I am a bit confused and would like any words of advice.

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Are you sure the theorem you state is correct? I think you made an error copying it, since the right side does not contain S at all. –  sxd Sep 9 '12 at 13:27
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What if I give you $a = -2$ and $b = -1$? –  Cocopuffs Sep 9 '12 at 13:28
    
Are you sure about the statement of the theorem? I think it will be "$x \in S$ such that $x \in (a,b)$" instead of "$x \in \mathbb{R}$". –  Sayantan Sep 9 '12 at 13:30
    
My bad. "A set $\mathbb{S}$ of real numbers is dense in $\mathbb{R}$ if every interval $I = (a,b)$ where $a < b$ contains a member of $\mathbb{S}$." –  Arthur Collé Sep 9 '12 at 13:31
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@Augustus What is the midpoint between the points $a=-2$ and $b=-1$ suggested by Cocopuffs? Also, constructing sentences is proof writing. The mathematical symbols are abbreviations for the words and sentences of the proof. I refuse to read my students proofs if there are more symbols than words. –  Kris Williams Sep 9 '12 at 13:35
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3 Answers

up vote 10 down vote accepted

That set is not dense in $\mathbb{R}$. Try picking an interval which only consists of negative numbers, does this interval contain any elements of $S$?

Finding such a counter example is enough to prove that $S$ is not dense in $\mathbb{R}$.

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Wow I am so dumb. Thank you. How does the density of a set relate to "boundedness" (having an upper and/or lower bound, and consequently having a sup or inf) ? –  Arthur Collé Sep 9 '12 at 13:39
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@Augustus: Clearly it is impossible for a bounded set to be dense in $\mathbb R$. But not all unbounded sets are dense -- for example $\mathbb Z$ is not dense. –  Henning Makholm Sep 9 '12 at 13:43
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@Augustus, Its not dumb at all, sometimes you get so fixed on a problem that you look so far for a solution while its right under your nose! –  sxd Sep 9 '12 at 13:51
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To give a third point of view:

A subset $D$ of a topological space $X$ is dense in $X$ if and only if $\overline{D} = X$ where $\overline{D}$ denotes the closure of $D$.

We also know that a set $C$ is closed if and only if $\overline{C} = C$.

But your set $S$ is closed and a proper subset of $\mathbb R$, hence cannot be dense in $\mathbb R$.

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Just to point out that mine and Matt's definitions are equivalent. Given a set $S$, the closure of $S$ is the set $\overline{S}$ consisting of $S$ together with all of its limit points. If $S$ is dense in $X$ then for all $x \in X$ we have $x \in \overline{S}$. Conversely, if $x \in \overline{S}$ for all $x \in X$ then $S$ is dense in $X$. –  Fly by Night Sep 9 '12 at 15:47
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In general topology, a subset $S$ of a topological space $X$ is called dense if every point of $X$ either belongs to $S$ or is a limit point of $S$. (A point $x \in X$ is a limit point of $S$ if every open neighbourhood of $x$ contains at least one point of $S$ different to $x$).

The non-negative real numbers are not dense in $\mathbb{R}$, under the metric topology, because, for example, $-1$ does not belong $S$, and $-1$ is not a limit point of $S$. (The open neighbourhood $\left(-\frac{3}{2},-\frac{1}{2}\right)$ contains $-1$ but does not contain any point of $S$.)

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