Is the interior of the union of $n$ closed balls equal to the union of the interiors of the $n$ closed balls? I am reading "Calculus on Manifolds" by Michael Spivak.
I am solving problem 1-22 on p.10 now.
If the following equality holds, I can solve the problem.
Let $B_1, \dots, B_n$ be closed balls in $\mathbb{R}^m$.
Intuitively, I guess the following equality holds, but I cannot prove that.
Does the following equality hold?
$$\operatorname{Int}(B_1 \cup \dots \cup B_n) = \operatorname{Int}(B_1) \cup \dots \cup \operatorname{Int}(B_n)$$

Here's a proof for the direction that seems to be correct. For a point P, if $P\in B_{Q_1}(\varepsilon_{Q_1}) \;\;\cup\;...\cup\;B_{Q_m}(\varepsilon_{Q_m})$, it also lies in at least one of the open balls, say $P\in B_{Q_j}(\varepsilon_{Q_j})$. Hence P lies in the interior of $\overline{B_{Q_1}(\varepsilon_{Q_1})}\;\cup \;...\cup \;\overline{B_{Q_m}(\varepsilon_{Q_m})}$, and we then have $$B_{Q_1}(\varepsilon_{Q_1}) \;\;\cup\;...\cup\;B_{Q_m}(\varepsilon_{Q_m})\subseteq int(\overline{B_{Q_1}(\varepsilon_{Q_1})}\;\cup \;...\cup \;\overline{B_{Q_m}(\varepsilon_{Q_m})})$$.
The other direction fails as demonstrated by counterexample in Bernstein's answer.
