Is $\mathbb N$ dense in $\mathbb R$? Is $\mathbb N$ dense in $\mathbb R$?
Let say $(a,b)=(0,1)$ How come it does not contain an element of $\mathbb N$?
Is it because it's an open interval?
Can someone please explain this to me? Also when would it be dense?
 A: $$\mathbb N = \{1,2,3,4,5,\ldots\}.$$
In order that $\mathbb N$ be dense in $\mathbb R$ it would be necessary for every open interval about every member of $\mathbb R$ to contain a member of $\mathbb N$.  Even if $(0,1)$ did contain a member of $\mathbb N$ (and it does not) there would still be open intervals about $0.5$ containing no member of $\mathbb N$, for example $(0.4,0.6)$.  The existence of at least one open interval containing $0.5$ and not intersecting $\mathbb N$ means $0.5$ is not a limit point of $\mathbb N$.  Only if every point in $\mathbb R$ were a limit point of $\mathbb N$ would $\mathbb N$ be dense in $\mathbb R$.
A: Every other answer tells you that in the standard topology of $\mathbb{R}$, $\mathbb{N}$ is not dense.
So, to get $\mathbb{N}$ in $\mathbb{R}$, you have to change the topology. Notice that $\mathbb{N}$ is indeed dense in $\mathbb{R}$ in the indiscrete topology (every non empty set is open in the indiscrete topology.)
A: Show that a subspace $\;D\;$ of a topological space $\;X\;$ is dense in it iff $\;D\cap Y\neq\emptyset\;$ for all open non-empty $\;Y\subset X\;$ .
Then, and under the usual topology of $\;\Bbb R\;$, your example shows $\;\Bbb N\;$ isn't dense here since $\;(0,1)\cap\Bbb N=\emptyset\;$
A: I'm going to assume you are using the standard topology on $\mathbb{R}$ (or $\mathbb{R}_{std}$ as I was taught). A subset of $A \subseteq \mathbb{R}$ is dense if $\bar A = \mathbb{R}$. Since $\mathbb{N}$ is closed in $\mathbb{R}$, then $\mathbb{N}=\mathbb{\bar N} \neq \mathbb{R}$. So no, $\mathbb{N}$ is not dense in $\mathbb{R.}$ If you need to prove that $\mathbb{N}$ is closed under $\mathbb{R}_{std}$, it suffices to observe that $$\bigcup _{i=1}^\infty (i,i+1) \cup(-\infty,1)$$ is open, and $$\left(\bigcup _{i=0}^\infty (i,i+1) \cup(-\infty,0) \right)^{c}=\mathbb{N}$$ is therefore closed.
A: The interval $(0,1)$ does not contain a natural number.
The fact doesn't follow from the fact that it is an open interval (after all, $(0,2)$ is also an open interval, and it contains $1$, which is a natural number).
Of course, there are natural numbers in the closed interval $[0,1]$ (namely $0$ and $1$, or just $1$ if you don't consider $0$ to be a natural number), but that has nothing to do with the closedness: the interval $[\frac{1}{3},\frac 2 3 ]$ is also closed, but doesn't contain any natural numbers.
However, it can be easily seen from the fact that $(0,1)$ is, by the definition, the set of real numbers strictly larger than $0$ and strictly smaller than $1$. But there are no natural numbers with that property, so there are no natural numbers in $(0,1)$.
Because $(0,1)$ is an open set, it intersects any dense subset of ${\bf R}$. This implies that ${\bf N}$ is not dense in ${\bf R}$, as it does not intersect $(0,1)$.
