A closed ball in a metric space is a closed set

Prove that a closed ball in a metric space is a closed set

My attempt: Suppose $D(x_0, r)$ is a closed ball. We show that $X \setminus D $ is open. In other words, we need to find an open ball contained in $X \setminus D$.

Pick $$t \in X-D \implies d(t,x_0) > r \implies d(t,x_0) - r > 0 $$ Let $B(y, r_1)$ be an open ball, and pick $z \in B(y,r_1)$. Then, we must have $d(y,z) < r_1 $. We need to choose $r_1$ so that $d(z,x_0) > r$. Notice by the triangle inequality

$$ d(x_0,t) \leq d(x_0,z) + d(z,t) \implies d(z,x_0) \geq d(x_0,t) - d(z,t) > d(x_0,t) - r_1.$$

Notice, if we pick $r_1 = d(t,x_0)-r$ then we are done.

Is this correct?

Solution 1:

Your proof consists of some correct steps done in the wrong order, which makes it something other than a valid proof. It looks more like scratchwork done in preparation for a proof. I rewrite it below, with some of the more important additions in bold. I will also change $t$ to $y$ throughout; when you wrote "$y$" you probably meant the same thing as "$t$".

Suppose $D(x_0, r)$ is a closed ball. We show that $X\setminus D(x_0,r) $ is open. In other words, for every point $y\in X\setminus D(x_0,r)$ we need to find an open ball contained in $X \setminus D$ with center $y$.

Since $y \in X\setminus D(x_0,r)$, it follows that $d(y,x_0) > r$, so $d(y,x_0) - r > 0 $. Let $r_1 = d(y,x_0)-r$.

I claim that the open ball $B(y, r_1)$ is contained in $X\setminus D(x_0,r)$. To prove this, consider any $z \in B(y,r_1)$. Notice by the triangle inequality

$$ d(x_0,y) \leq d(x_0,z) + d(z,y) \implies d(z,x_0) \geq d(x_0,y) - d(z,y) > d(x_0,y) - r_1 =r.$$ This shows $z\in X\setminus D(x_0,r)$, which completes the proof.