# Thread: Metric spaces, open sets, and closed sets

1. ## Metric spaces, open sets, and closed sets

I have these two propositions stated in my book as follows.

Proposition 1.
Let $\displaystyle (X,d)$ be a metric space, then
a) The sets $\displaystyle X$ and $\displaystyle \emptyset$ are open
b) Any finite intersection of open sets is open
c) Any union of open sets is open.

Similarly,
Proposition 2.
Let $\displaystyle (X,d)$ be a metric space, then
a) The sets $\displaystyle X$ and $\displaystyle \emptyset$ are closed.
b) Any finite union of closed sets is closed
c) Any intersection of closed sets is closed.

I understand parts b) and c) of both propositions. However, I have no clue why a) is listed for both and why a) is important. What is a) telling us? If someone could please explain to me the meaning of a) in both cases it would be much appreciate. Thank you very much.

2. Originally Posted by Sheld
Proposition 1.
Let $\displaystyle (X,d)$ be a metric space, then
a) The sets $\displaystyle X$ and $\displaystyle \emptyset$ are open
Similarly,
Proposition 2.
Let $\displaystyle (X,d)$ be a metric space, then
a) The sets $\displaystyle X$ and $\displaystyle \emptyset$ are closed.
However, I have no clue why a) is listed for both and why a) is important. What is a) telling us? If someone could please explain to me the meaning of a) in both cases it would be much appreciate.
A set is open if and only if its complement is closed.
A set is closed if and only if its complement is open.

What is the complement of $\displaystyle X~?$

What is the complement of $\displaystyle \emptyset~?$

3. Definition of "open set"- A is open if and only if, "If x is in A, then there exist d> 0 such that N(x,d), the set of all points whose distance from x is less than d, is a subset of A.

If A is itself the set of all points, what does that tell you? If A is empty what happens to the hypothesis "if x is in A"?

4. So we can prove directly that $\displaystyle X$ and $\displaystyle \emptyset$ are open?

Therefore we know that $\displaystyle X^{C} = \emptyset$ and $\displaystyle \emptyset^{C} = X$are closed?

5. Originally Posted by sheld
so we can prove directly that $\displaystyle x$ and $\displaystyle \emptyset$ are open?
Therefore we know that $\displaystyle x^{c} = \emptyset$ and $\displaystyle \emptyset^{c} = x$are closed?
exactly!