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.


4 Answers 4


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.

$\quad\quad\quad\quad\quad\quad\quad$enter image description here

  • 1
    $\begingroup$ Really great answer thanks, how did you make the diagram? What software did you use? $\endgroup$ Dec 20, 2014 at 12:39
  • 3
    $\begingroup$ @Dansmith Thanks. The software was JSXGraph. $\endgroup$ 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .