An open ball is an open set

Prove that for any $x_0 \in X$ and any $r>0$, the open ball $B_r(x_o)$ is open.

My attempt: Let $y\in B_r(x_0)$. By definition, $d(y,x_0)<r$. I want to show there exists an $r_1\in\mathbb{R^+}$ s.t. $B_{r_{1}}(y)\subseteq B_r(x_0)$. Let $a\in B_{r_{1}}(y)$. Then, $d(a,y)<r_1$. For $a\in B_{r}(x_0)$, $d(a,x_0)<r$. I want to show $d(a,y)<r_1$ implies $d(a,x_0)<r$. By triangle inequality, $d(a,y)\leq d(a,x_0) + d(y,x_0) \rightarrow$ $d(a,y)<r_1\leq d(a,x_0)+d(y,x_0)<2r...$

I'm a little stuck after this point.

You need to specify $$r_1$$.

For $$y\in B_r(x_0)$$, let $$r_1=r-d(y,x_0)$$. Then if $$x\in B_{r_1}(y)$$ $$d(x,x_0)\le d(x,y)+d(y,x_0)< r_1+ d(y,x_0)=r.$$ So $$B_{r_1}(y)\subseteq B_r(x_0)$$. This shows that $$B_r(x_0)$$ is open.

$$\quad\quad\quad\quad\quad\quad\quad$$

• Really great answer thanks, how did you make the diagram? What software did you use? Commented Dec 20, 2014 at 12:39
• @Dansmith Thanks. The software was JSXGraph. Commented Dec 20, 2014 at 12:46

Since you want to conclude that $d(a,x_0)$ isn’t too big, you should apply the triangle inequality to it, not to $d(a,y)$: $d(a,x_0)\le d(a,y)+d(y,x_0)<r_1+d(y,x_0)$. In other words, you need to choose $r_1$ so that $r_1+d(y,x_0)\le r$. Can you explain now how to do that? Remember, $y$ is a particular point of $B_r(x_0)$, so $d(y,x_0)$ is a fixed number.

In the second line, you have indicated that you want to find an $r_1$ such that $d(a,y)<r_1$. This is good.

First, what can you say about $d(x_0,y)$, since $y$ is in $B_r(x_0)$? $d(x_0,y)=p \leq r$.

Now, the choice for $r_1$ is clearer since we want $B_{r_1}(y)$ to be inside $B_r(x_0)$. Let it be $r_1=r-p$, since this is intuitively how much farther from $x_0$ our points are allowed to lie while still being in the ball $B_r(x_0)$.

Finally, lets show that this was a good choice. Pick a point $a$ in $B_{r_1}(y)$ Then by the triangle inequality focusing on $d(x_0,a)$, we get $d(x_0,a) \leq d(x_0,y)+d(y,a)$, and since we so cleverly chose our $r_1$, things will work out nicely. It is vital that you focus on $d(x_0,a)$. You were focusing instead on $d(x_0,y)$ which is why you did not get you the bound you want. This will show that $a$ is indeed in the open ball $B_r(x_0)$.

Let $B_r^{\delta}(a)$ be given and $x \in B_r^{\delta}(a)$ then $\delta(a, x) = \epsilon < r$

now look at $B_{\gamma}^{\delta}(x)$ for $\gamma < r - \epsilon$ then $\forall y \in B_{\gamma}^{\delta}(x)$ we have $\delta(y,x) < \gamma$

so then $\delta(y,a) <= \delta (y,x) + \delta(x,a) <= r - \epsilon + \epsilon = r$ and the Result follows.