Real Analysis: Is $\emptyset$ a open set in $X$?

Question

I'm learning real analysis.

A subset $G$ of $X$ is called open if for each $x \in G$ there is a neighborhood of $x$ that is contained in G

My question is that is $\emptyset$ a open set in $X$?

The set $\emptyset$ has no elements, so there is no neighborhood of $x$ is contained in $G$. Hence, it is not open set. However my intuitive tell me it should be a open set. What's wrong? Could anyone explain it? Thanks.

It is open, since the implication is trivially (vacuously) satisfied. — Tyler, Jan 10 '13 at 03:50
See Why is predicate “all” as in all(SET) true if the SET is empty? — , Jan 10 '13 at 04:05
Also, when you say "there is no neighborhood of $x$ [that] is contained in $G$", what do you mean by $x$? — , Jan 10 '13 at 04:08
The phrase "there is no neighborhood of $x$ is contained in $G$" is neither true nor false: instead it is a predicate whose domain is the empty set. It would yield a particular truth value if you plugged in an element of the empty set, but since there aren't any, you can't even do that! — , Jan 10 '13 at 04:08

score 3 · Answer 1 · answered Jan 10 '13 at 03:55

A subset of a metric space $X$ is either open or not open. If $\emptyset$ were not open, there would be a point $x \in \emptyset$ such that there exists no neighborhood of $x$ contained in $\emptyset$. However, by definition there are no points at all in $\emptyset$. Hence, $\emptyset$ is open.

score 2 · Answer 2 · answered Jan 10 '13 at 03:51

2

In general, for any statement $P$, for every element of $\emptyset$, the statement $P$ is vacuously true.

answered Jan 10 '13 at 03:51

score 2 · Answer 3 · answered Jan 10 '13 at 03:57

2

Like you said:

A subset G of X is called open if for each x∈G there is a neighborhood of x that is contained in G

If G is empty you can not choose a x that not satisfies this. So G is open.

answered Jan 10 '13 at 03:57

If G is empty you can not choose a x that satisfies this. So G is not open. – Jill Clover Jan 10 '13 at 04:02
@TengPeng I wanted put NOT – Jan 10 '13 at 04:04
@Teng: Openness is not "there exists an $x$ in $G$ such that $x$ has a neighborhood contained in $G$". – Jan 10 '13 at 04:39

Real Analysis: Is $\emptyset$ a open set in $X$?

3 Answers3