There is this small "trivial" part of a proof that really bugs me?

Question

I am trying to show: If $A=B_r(x)\subset X$ where $(X,d)$ is an arbitrary metric space (and $B_r(x)$ is an open ball of radius $r$ centred at $x$), then int $A=A$.

int $A\subset A$ is obvious. I am having a difficulty showing the other inclusion.

Let's say $y\in A$, then $d(y,x)=s$ (say) $<r$.

Here is my question: Is it true that "If $z\in B_s(y)$ then $d(z,y)<r-s$" ? If yes, how can we show it?

We all know that $d(z,y)<s<r$, but how can we show that $d(z,y)<r-s$" ?

That's the part that bugs me. If that statement is true, I can finish the rest of the proof.

Many thanks!

You're not obligated to consider $B_s(y)$. In fact, if you considered $B_{r-s}(y)$... — Joey Zou, Oct 27 '15 at 00:58
No, why would that be true? For example, if $X=\mathbb R$ with the usual metric, and $x=0,r=3$ and $y=2$. Then if $|z-y|<2$ is $|z-y|<3-2$? — Thomas Andrews, Oct 27 '15 at 00:59
No, it's not true. Draw a picture, and I think you'll see what ball you do want to use. — Nate Eldredge, Oct 27 '15 at 01:00

score 1 · Accepted Answer · answered Oct 27 '15 at 01:00

Saying that ${\rm int}\, A = A$ is the same as saying $A$ is open, or that every point of $A$ is an interior point. If you pick $z\in B_r(x)$, let $s=d(x,z)<r$. A drawing should help you prove that a ball of radius less or equal to $r-s>0$ centered at $z$ is contained in $B_r(x)$. Indeed, suppose that $d(z,y)<r-s$. Then

$$d(y,x)\leqslant d(y,z)+d(x,z)< r-s+s=r$$

So that $y\in B_r(x)$.

There is this small "trivial" part of a proof that really bugs me?

1 Answers1