"In the space of rational numbers with the usual topology (the subspace topology of $\Bbb R$), the boundary of $(-\infty,a)$, where a is irrational, is empty."
There are many ways to find boundary points like $A \subseteq X$:
- $\operatorname{cl}(A)\setminus \operatorname{int}(A)$
- $\operatorname{cl}(A)\cap \operatorname{cl}(X\setminus A)$
But i am using the definition that a point is a boundary point if its every open neighborhood contains some points of the set and some points outside of the set.
I feel like $a$ in $(-\infty, a)$ should be boundary point as any open neighborhood ( by subspace topology open sets will be $(a-\epsilon,a+\epsilon) \cap \Bbb Q$ of $a$ will contain some points of $(-\infty, a)$ and some outside of it. But i am not getting where I am wrong as they say that the set has no boundary point.
Best Answer
$a$ can only be a boundary point of $A=(-\infty, a)$ in $\Bbb Q$ when $a$ is a member of $\Bbb Q$ in the first place! The set $A$ is a set of rationals (!) and its closure and interior points are also subsets of the rationals, because that is the "universe" we're considering. So you cannot consider neighbourhoods of a point, that's not even in your space at all!
The "mean" thing here is that the set is defined in terms of a point $a$ that is not in $\Bbb Q$ but in $\Bbb R\setminus \Bbb Q$. But this makes $A$ closed and open at the same time in $\Bbb Q$ because $\operatorname{cl}_{\Bbb Q}(A)= \operatorname{cl}_{\Bbb R}(A) \cap \Bbb Q = (-\infty,a] \cap \Bbb Q = A$.
So $\partial A= \operatorname{cl}(A)\setminus \operatorname{int}(A)=A\setminus A=\emptyset$.