# Thread: Closure (Metric Space)

1. ## Closure (Metric Space)

Let A, B be subsets of a metric space. Show that $\displaystyle \overline{A \cup B} = \bar{A} \cup \bar{B}$ and that $\displaystyle \overline {A \cap B} \subset \bar{A} \cap \bar{B}$. Give an example to show that $\displaystyle \overline {A \cap B}$ and $\displaystyle \bar{A} \cap \bar{B}$ may not be equal.

2. Originally Posted by aliceinwonderland
Let A, B be subsets of a metric space. Show that $\displaystyle \overline{A \cup B} = \bar{A} \cup \bar{B}$ and that $\displaystyle \overline {A \cap B} \subset \bar{A} \cap \bar{B}$.
What are you having trouble with?
Give an example to show that $\displaystyle \overline {A \cap B}$ and $\displaystyle \bar{A} \cap \bar{B}$ may not be equal.
Consider when $\displaystyle A,B$ are seperated but share a limit point. Example $\displaystyle A=(-1,0),B=(0,1)$ because $\displaystyle A\cap B=\varnothing\implies\overline{A\cap B}=\varnothing$ but $\displaystyle 0$ is a limit point of both so $\displaystyle \bar{A}\cap\bar{B}=\left\{0\right\}$

3. Originally Posted by aliceinwonderland
Let A, B be subsets of a metric space. Show that $\displaystyle \overline{A \cup B} = \overline{A} \cup \overline{B}$ and that $\displaystyle \overline {A \cap B} \subset \overline{A} \cap \overline{B}$. Give an example to show that $\displaystyle \overline{A \cap B}$ and $\displaystyle \overline{A} \cap \overline{B}$ may not be equal.

a) One inclusion: $\displaystyle \overline{A} \cup \overline{B}$ is closed since it is the finite union of closed sets and it contains $\displaystyle A \cup B$ so it must contain the closure $\displaystyle \overline {A \cup B}$.

Other inclusion: STS $\displaystyle \cup \overline{A_\alpha} \subseteq \overline{\cup A_\alpha}.$ Well, $\displaystyle \forall \alpha, A_\alpha \subseteq \overline{\cup A_\alpha}.$ So $\displaystyle \overline{A_\alpha} \subseteq \overline{\cup A_\alpha}$. Therefore $\displaystyle \cup \overline{A_\alpha} \subseteq \overline{\cup A_\alpha}.$

Next Part: $\displaystyle A \cap B$ is closed set containing $\displaystyle A \cap B$. So it must contain its closure $\displaystyle \overline{A \cap B}$

next part: take $\displaystyle A=(0,1)$ and $\displaystyle B=(-1,0)$ then $\displaystyle \overline{A \cap B} = \emptyset$ but $\displaystyle \bar{A} \cap \bar{B}=\{0\}$. QED