Closure in the Discrete Topology

Question

If $\tau$ is the discrete topology on the real numbers, find the closure of $(a,b)$

Here is the solution from the back of my book:

Since the discrete topology contains all subsets of $\Bbb{R}$, every subset of $\Bbb{R}$ is both open and closed. Therefore, the closure of $(a,b)$ is $[a,b]$.

Whaaat?! This must be a mistake. Please tell me this is a mistake.

Answer 1

The solution is wrong. Every set in the discrete topology is both open and closed. Closure of a closed set is itself. Hence, the answer should be $(a,b).$

Answer 2

The above comments and answers are absolutely correct.

But we can prove that there is a mistake in your book,by contradiction.

Suppose that $cl(a,b)=[a,b]$ Then $a \in cl(a,b)=[a,b]$ and form definition of closure,we know that a point $x$ is in the closure of a set $A$ in a metric space $X$ if every open ball with center $x$ intersects the set $A$

Now for $A=(a,b)$ we have that the ball $B(a,\frac{1}{2})=\{a\}$ in a discrete metric space $(\mathbb{R},d_{dis})$ and $\{a\} \cap (a,b)= \emptyset$

We contradict the definition of closure.