If $(K_i)_{i \in \mathbb{N}}$ is a sequence of closed sets in $\mathbb{R}^3$, then the union of these sets $\bigcup_{i=1}^\infty K_i = K_1 \cup K_2 \cup … $ is also closed.
My idea: ($\bigcup_{i=1}^\infty K_i)^C = \bigcap_{i=1}^\infty (K_i)^C$
I.e. , that if all $K_i$ are open, then $\bigcap_{i=1}^\infty K_i = K_1 \cap K_2 \cap..$ is open, which is wrong.
Counterexample would be
$$K_i := \{(x,y,z) \in \mathbb{R^3}: \Big\| \begin {pmatrix} x \\ y \\ z \end{pmatrix} \Big\| < \frac{1}{i} \}$$
Is this correct?
Best Answer
The above statement is, as you suggest, false. Let $(K_i)_{i \in \mathbb{N}}$ be a sequence of closed sets in $\mathbb{R^n}$. Then $\cup_{i=1}^\infty K_i$ is closed iff $(\cup_{i=1}^\infty K_i)^C$ is open, that is iff $\cap_{i=1}^\infty K_i^C$ is open.
Note that each $U_i:=K_i^C$ is open, and as you suggest, you could choose the $U_i$'s so that their intersection is not open, e.g. $U_i = \{v \in \mathbb{R^n} \mid \|v\| < \frac{1}{i}\}$, $\cap_{i=1}^\infty U_i=\{0\}$.
As others have suggested, you could alternatively give a counterexample for the original statement by choosing e.g. $K_i=\{v \in \mathbb{R}^n \mid \|v\| \leq 1-\frac{1}{i}\}$ so that $\cup_{i=1}^\infty K_i=\{v \in \mathbb{R}^n \mid \|v\| < 1\}$.
P.S. In your question you used the same symbol $K_i$ to stand for two different notions (a closed and an open set). This is a very bad practice which you should avoid as it confuses the reader.