A+B的閉包不一定也是A的閉包+B的閉包

examples of the closure of A+B is not included in the clousre of A plus the closure of B

in R \mathbb{R} R

1. Let A = { n + 1 n : n ∈ Z + } A=\{n+\frac{1}{n}: n\in\mathbb{Z}^{+}\} A={n+n1​:n∈Z+} and B = { − n : n ∈ Z + } B=\{-n:n\in \mathbb{Z}^{+}\} B={−n:n∈Z+}
2. Clearly, A A A and B B B are closed. Hence, A = A ‾ A=\overline{A} A=A and B = B ‾ B=\overline{B} B=B.
3. 0 ∉ A ‾ + B ‾ 0\notin \overline{A}+\overline{B} 0∈/A+B.
4. However, { 1 n : n ∈ Z + } ⊂ A + B \{\frac{1}{n}:n\in \mathbb{Z}^{+}\}\subset A+B {n1​:n∈Z+}⊂A+B, so 0 ∈ A + B ‾ 0\in \overline{A+B} 0∈A+B​.
5. Step 3 and step 4 tells that A + B ‾ \overline{A+B} A+B​ is not a subset of A ‾ + B ‾ \overline{A}+\overline{B} A+B.

in R 2 \mathbb{R}^2 R2

1. Let A = { ( x , 1 / x ) : x ≠ 0 } A=\{(x,1/x): x\ne 0\} A={(x,1/x):x=0} and B = { ( x , − 1 / x ) : x ≠ 0 } B=\{(x,-1/x): x\ne 0\} B={(x,−1/x):x=0}.

   A: x ∈ [ − 10 , 10 ] x\in [-10,10] x∈[−10,10], step size 1 1 1

   

   B: x ∈ [ − 10 , 10 ] x\in [-10,10] x∈[−10,10], step size 1 1 1

1. Clearly, A A A and B B B are closed.
2. Then A + B A+B A+B can be illustrated as