# Thread: Verifying the Boundaries/Interiors etc of these two functions

1. ## Verifying the Boundaries/Interiors etc of these two functions

Hi, I believe I have the right information for the following two functions but I just need a brief statement as to why it is true.

Consider the set $\displaystyle \mathbb{Q}$ of rational numbers.

1) $\displaystyle \partial(Interior(\mathbb{Q}))$, which I said is $\displaystyle \varnothing$

2) $\displaystyle \partial((Interior(\mathbb{Q})^C))$, which I said is $\displaystyle \varnothing$

3) $\displaystyle Closure(Interior(\mathbb{Q}))$, which I said is $\displaystyle \varnothing$

And for the second function, S={(x,y): y = sin(1/x), x>0} I said the following:

1) $\displaystyle Interior(S)$, which I said is $\displaystyle \varnothing$

2) $\displaystyle \partial(S)$, which I said is $\displaystyle \{(x,y): x=0, -1 \le \right y \le \right 1\}$

3) $\displaystyle Closure(S)$, which I said is $\displaystyle S \cup \{(x,y): x=0, -1 \le \right y \le \right 1\}$

Again, I just need brief statements verifying the validity of these statements using point/ball methods used in elementary topology. Thanks a lot, I appreciate it.

2. Yes, those are all true. But why do you think they are true? That's what you want to say!

In R, a "ball" is an interval. And any interval contains both rational and irrational numbers.

In $\displaystyle R^2$, a "ball" is a disk. All you need to say is that a disk on any point of a curve must contain some points not on the curve.

3. Actually $\displaystyle \partial(S)=\text{closure}(S)$.
Every point in $\displaystyle S$ is boundary point of $\displaystyle S$.
Any open disk containing a point of $\displaystyle S$ must contain a point not in $\displaystyle S$.