I'm trying to understand the definition of open sets and interior points in a metric space.
I'm not sure why $$ Int(\mathbb{N})=\phi.$$
$ A\subseteq X, a \in X $, then $a$ is said to be an Interior Point of $A$ if $ \exists r \in \mathbb{R} >0 $ such that $ U(a,r) \subseteq A. $
Where $U(a,r)$ denotes the open ball center $a$ radius $r$.
Now applying this definition letting $X= \mathbb{R}$ and $A=\mathbb{N}$ and the usual metric on $\mathbb{R}$, then could I not take $a=2, r=3$ then take say $x=1$, then $d(a,x)=d(2,1)=|2-1|= 1 < 3.$ Clearly $1 \in \mathbb{N}$…
Or have I misunderstood and it's the case that every point in the open ball has to belong to $A$?
Thanks.
Best Answer
No, your reasoning doesn't work.
If, as you say, you take $a=2, r=3$ you have to ask whether $U(2,3)\subseteq \mathbb N$.
But $U(2,3)$ is the open interval $(-1,5)$, which is not a subset of $\mathbb N$. For example, $(-1,5)$ contains the point $\frac12$ which is not in $\mathbb N$.