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

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.

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It is open, since the implication is trivially (vacuously) satisfied. – Tyler Jan 10 '13 at 3:50
– Rahul Jan 10 '13 at 4:05
Also, when you say "there is no neighborhood of $x$ [that] is contained in $G$", what do you mean by $x$? – Rahul Jan 10 '13 at 4:08
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! – Hurkyl Jan 10 '13 at 4:08

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.

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

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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.

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If G is empty you can not choose a x that satisfies this. So G is not open. – John Hass Jan 10 '13 at 4:02
@TengPeng I wanted put NOT – user52188 Jan 10 '13 at 4:04
@Teng: Openness is not "there exists an $x$ in $G$ such that $x$ has a neighborhood contained in $G$". – Hurkyl Jan 10 '13 at 4:39