# How to prove properties of the family of closed sets in a metric space

I know that is true:

Let $(X, d)$ a metric space. The family $\mathcal {U}$ of all open subsets of $X$ has these properties:

$1)$ $\phi, X\in \mathcal {U}$;

$2)$ $U_1, U_2 \in \mathcal {U}\Rightarrow U_1\cap U_2 \in \mathcal {U}$;

$3)$ $\{U_i\}_{i\in I}\subset\mathcal{U}\Rightarrow \bigcup_{i\in I} U_i\in\mathcal{U}$

Prove: $1)$ it is clear (because $\phi$, is a subset for every set $\Rightarrow\phi \in\mathcal {U}$, every set is a subset of its own $\Rightarrow X \in\mathcal {U};$ If I'm wrong please correct)

$2)$ Let $x\in U_1\cap U_2.$ Then exist $r_1, r_2>0$ such that $T(x,r_i)\subset U_i, i=1,2.$ Let $r=\min\{r_1,r_2\}.$ Then $r>0$ and $T(x,r)\subset U_1\cap U_2.$

$3)$ Let $\{U_i\}_{i\in I}\subset\mathcal{U} \text{and} x\in\bigcup_{i\in I} U_i,$ then $x\in U_{i_0}$ for some $i_0\in I$. For opennes of the $U_{i_0}$ (why is $U_{i_0}$ open, please tell me), exist $r>0$ such that $T(x,r)\subset U_{i_0}\subset \bigcup_{i\in I} U_i$

But I didnt now how to prove this theorem:

Let $(X, d)$ a metric space. The family $\mathcal {F}$ of all closed subsets of $X$ has these properties:

$1)$ $\phi, X\in \mathcal {F}$;

$2)$ $F_1, F_2 \in \mathcal {F}\Rightarrow F_1\cup F_2 \in \mathcal {F}$;

$3)$ $\{F_i\}_{i\in I}\subset\mathcal{F}\Rightarrow \bigcap_{i\in I} F_i\in\mathcal{F}$

I now that using a formula of DeMorgan but I didnt now how to should do. Help me plase to prove. Thanks for your help and your attention.

• Please replace the title of this question by something meaningful. Commented Nov 3, 2013 at 10:43
• just did a thing the mr, thanks @Brian M. Scott Commented Nov 3, 2013 at 10:47

Atahualpa has answered the main part of your question; here are some comments on the first part.

I’m assuming that your definition of open set is that $U$ is open if and only if for each $x\in U$ there is an $r_x>0$ such that $T(x,r_x)\subseteq U$, where $T(x,r_x)$ is the open ball of radius $r_x$ centred at $x$.

1. $X$ is open because for each $x\in X$, $T(x,1)\subseteq X$. $\varnothing$ is open because it’s vacuously true that $T(x,1)\subseteq\varnothing$ for each $x\in\varnothing$, since there is no $x\in\varnothing$.

3. What you’re proving here is that if $\{U_i:i\in I\}$ is a family of open sets, then $\bigcup_{i\in I}U_i$ is also open. Thus, $U_{i_0}$ is open by hypothesis.

• Thanky very much sir, I knew very well Commented Nov 3, 2013 at 10:56
• @Madrit: You’re welcome. Commented Nov 3, 2013 at 10:57
• Mr. has set of which the $R$ is also open and closed Commented Nov 3, 2013 at 10:58
• @Madrit: If $\langle X,d\rangle$ is a metric space, the sets $\varnothing$ and $X$ are always both open and closed. Commented Nov 3, 2013 at 10:59
• that is to say can be said that set $phi$ is at the same time also open the closed, or not Commented Nov 3, 2013 at 11:01

You know that $F$ is closed in $X$ iff $X \setminus F$ is open.

So, to show that if $F_1, F_2$ are closed, then $F_1 \cup F_2$ is closed, is just routine. For instance. $X \setminus F_1, X \setminus F_2$ are open, hence by first part,

$$(X \setminus F_1 ) \cap (X \setminus F_2 ) = X \setminus (F_1 \cup F_2) \; \; \text{is open}$$

Hence, $F_1 \cup F_2$ must be closed, that i $F_1 \cup F_2 \in \mathcal{F}$.

Similarly if $\{ F_i \}$ are closed subset of $X$, then $\{ X \setminus F_i \}$ are open.

$$\therefore \bigcup X\setminus F_i = X \setminus \bigcap F_i \; \; \text{is open}$$

Hence, $\bigcap F_i$ is closed. In other words, $\bigcap F_i \in \mathcal{F}$

• it is very clear answer, thanks sir, but why $\phi, X\in U$ or why $\phi, X\in F$ I didnt now, and why $U_{i_0}$ is open, thanks Commented Nov 3, 2013 at 10:50
• $X, \varnothing$ are both open and closed Commented Nov 3, 2013 at 10:51
• $U_{i_0}$ is open because if is a member of a the collection of open sets $\{ U_i \}$ Commented Nov 3, 2013 at 10:52
• if the required set of $R$ which is also open or closed can say that it is its metric space or should we say that the set of open and closed in $R$ is $\phi$ Commented Nov 3, 2013 at 10:53