# closure

• Apr 26th 2009, 06:05 PM
xalk
closure
prove:

$\displaystyle \overline\emptyset = \emptyset$

$\displaystyle \overline X = X$
• Apr 26th 2009, 10:49 PM
Gamma
both sets are both open (definition of topology). In particular, they are complements of one another, thus the complement of each is open and hence each one is also closed. The closure of a closed set is itself.
• Apr 27th 2009, 12:49 AM
xalk
THANK you,but i meant to prove the two relations by using the definition of a point of adherence and not the theorem:

A is closed iff $\displaystyle \overline A = A$
• Apr 27th 2009, 01:03 PM
Gamma
So you mean like $\displaystyle \overline A = A \cup A'$?

1) Suppose $\displaystyle x\in X$ were a "point of adherence" as you guys call them apparently. Then for any open U containing x, you need $\displaystyle U \cap \emptyset \not = \emptyset$ but this is clearly not possible. So the set of all points of adherence of $\displaystyle \emptyset$ is $\displaystyle \emptyset$.

2) Similarly suppose $\displaystyle x\in X$ were a "point of adherence." Then for any open U containing x, $\displaystyle U \cap X = U \supset \{x\} \not = \emptyset$. So every point in X is a point of adherence, so yeah, you are done.