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When I took calculus my teacher gave us a definition for a set $A$ to be dense in a set $B$:

Let $A,B \subset \mathbb R$. Then $A$ is dense in $B$ iff $\forall x\in B$ and $r>0$ $V^\circ_r(x)\cap A \neq \varnothing$ where $V^\circ_r(x)$ denotes the ball of radius $r$ around $x$ with the point $x$ removed, i.e., if $y\in V^\circ_r(x)$ then $0<|x-y|<r$.

So with this definition $\mathbb Q$ is dense in $\mathbb R$; $\mathbb Q$ is dense in $\mathbb Q$; but $\mathbb N$ is not dense in $\mathbb N$

I tried to generalize this definition in a topological space.

However in Brian M. Scott's answer here, there is another definition of a set being dense in another set, which applies to general topological spaces:

$A$ is dense in $B$ iff $B\subset \overline{A}$.

With this defintion $\mathbb N$ is dense in $\mathbb N$

So what is the correct definition? If it is the first one then what would be its generalization to a topological space?

Community
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user128422
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2 Answers2

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Firt of all, I'm assuming that the condition

$\forall x\in B$ and $r>0$ $V°_r(x)\cap A \neq \varnothing$

really means

for all $x\in B$ and for all $r>0$, $V^0_r(x)\cap A\neq\emptyset\qquad$ (1)

Note that (1) is equivalent to

every point in $B$ is a limit point of $A$.

This condition is stronger then $B\subset\overline{A}$ in general, which is difference between (1) and the one you quoted in the linked question. If $A$ is perfect, which means $A$ is closed and every point of $A$ is a limit point of $A$, then two definitions are the same.

In real analysis, I have never seen one uses (1) to define a dense set. On the other hand if one uses $V_r(x)=\{y\in\mathbb{R}\mid |y-x|<r\}$ instead of $V^0_r(x)$ in (1), then the two definitions are equivalent.

I'm quite curious why one would talk about dense sets in a calculus course and what textbook one is using.

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I think that the definition of densness in topological space is $A\subset B$ is dense in $B$ if $ \bar{A} =B$ by this definition $\mathbb{N}$ is not dense in any other set. An equivalent definition is :

$A$ is said to be dense in $B$ if every element of $B$ is limit of a sequence of elements of $A$.

As we know every real number is limit of a sequence of rational numbers. But what about $\mathbb{N}$?

Ali
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  • $A$ is dense in $ B$ iff ($A\subset B\land \bar A\supset B$). For example, in $\mathbb R,$ if A=$(0,1)\cap \mathbb Q$ and $B=(0,1)$ then $A$ is dense in $B,$ and $\bar A=[0,1].$ – DanielWainfleet Jan 07 '17 at 04:21