Metric spaces, open sets, and closed sets

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March 16th 2011, 12:04 PM
Sheld

Metric spaces, open sets, and closed sets

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

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

Similarly,
Proposition 2.
Let $(X,d)$ be a metric space, then
a) The sets $X$ and $\emptyset$ 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.
March 16th 2011, 12:25 PM
Plato

Quote:

Originally Posted by Sheld

Proposition 1.
Let $(X,d)$ be a metric space, then
a) The sets $X$ and $\emptyset$ are open
Similarly,
Proposition 2.
Let $(X,d)$ be a metric space, then
a) The sets $X$ and $\emptyset$ 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 $X~?$

What is the complement of $\emptyset~?$
March 16th 2011, 01:19 PM
HallsofIvy

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"?
March 16th 2011, 05:14 PM
Sheld

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

Therefore we know that $X^{C} = \emptyset$ and $\emptyset^{C} = X$ are closed?
March 16th 2011, 05:17 PM
Plato

Quote:

Originally Posted by sheld

so we can prove directly that $x$ and $\emptyset$ are open?
Therefore we know that $x^{c} = \emptyset$ and $\emptyset^{c} = x$ are closed?

exactly!