Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
It is open, since the implication is trivially (vacuously) satisfied. –  Tyler Jan 10 '13 at 3:50
    
    
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

3 Answers 3

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.

share|improve this answer

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

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.