Is $\emptyset$ in $R^n$ an open set?

Question

The complement of $\emptyset$ in $R^n$ is $R^n$ itself, which is an open set itself. So, $\emptyset$ should be a closed set, by definition of closed set. However, Tom M. Apostol's book says it is an open set. Am I going wrong somewhere in my argument? If so, where?

The empty set is both closed and open (same with $\mathbb{R}^n$) — Alvin Jin, Oct 03 '15 at 07:05
The symbol for the empty set is not Phi: $\phi ≠ \emptyset$ (\phi ≠ \emptyset). — k.stm, Oct 03 '15 at 07:09
Empty set is both open and closed set in any space $X$ regardless of it's topology! — Error 404, Oct 04 '15 at 10:50

score 4 · Answer 1 · answered Oct 03 '15 at 07:21

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Note that open and closed are not mutually exclusive descriptors (though you might be used to thinking that way due to learning about the closed $[a,b]$ and the open $(a,b)$ first). A set can be both open and closed (such as $\mathbb{R}^n$ and $\emptyset$).

answered Oct 03 '15 at 07:21

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Remind me this https://www.youtube.com/watch?v=SyD4p8_y8Kw – Oct 03 '15 at 07:56

score 3 · Answer 2 · answered Oct 03 '15 at 08:44

Sets can be open, closed, both or neither.

It's all about points. Consider a set A and consider points in and out of A.

If every neighborhood of a point x, whether or not x is in A or not, has a point (other than x) that is part of A, we call that a limit point. Intuitively it is right up next to some point of A. If a limit point isn't in A, then it's right up on the "edge" of A.

A is closed if all its limit points are in A. The empty set is closed because it doesn't have any limit points. (No point is right up next to the empty set because the empty set has no points to get next to.) R^n is closed because all points are in R^n all the point next to points in R^n are also in R^n.

If a point x is in A, if all the points within a small distance of x are also in A then we call x an interior point. Intuitively there is a fuzzy ball of A around the interior point x. (But this can be misleading because if your metric space isn't R^n it's possible that there are no points in your metric space within the distance from x. x would still be an interior point because x is the only point within the distance from x and x is in A.)

Interior points can be limit points but they don't have to be. (They usually are because all neighborhoods contain some of the fuzzy ball around the interior point and that ball contains some other points of A. But if this is a metric space with other points near the interior point then it isn't a limit point.) limit points can be interior points but they don't have to be. (A limit point could be that it's right on the "edge" of A. Every neighborhood has a point of A but also points not on A.) An interior point is always in A. A limit point may or may not be in A.

If every point in A is an interior point of A then we say A is an open set. (Intuitively this is like saying no point of A is on the "edge" of A.) The empty set is open because (ha!) all of its points are interior points. (Because it has no points! har-har.) R^n is open because all of the point or R^n have fuzzy balls entirely in R^n around them.

So. Sets can be open, closed, both or neither. If a set is open then its complement is closed. This is because if all points of A are interior points of A, then none of the points of A can be limit points of comp(A). (Because there are neighborhoods of that are entirely in A so these neighborhoods don't have any points of comp(A) so they can't be limit points of comp(A).) So all the limit points of comp(A) are in comp(A). So comp(A) is closed.

If a set is closed then its complement is open. This is because if A is closed all the limit points are in A so so none of the all the points of comp(A) are not limit points of A. This means its not the case that every neighborhood of a point in comp(A) has a point in A. This means for every point of comp(A) there is some neighborhood with no point in A. This means all the points in the neighborhood are in comp(A). All points of comp(A) have such a neighborhood so all points of comp(A) are interior point of comp(A). So comp(A) is open.

And that's it. open and closed are very closely related but not mutually exclusive The empty set is both.

score 1 · Answer 3 · answered Oct 03 '15 at 07:15

The empty set $\varnothing$ is open in $\mathbb{R}^{n}$ because it vacuously satisfies the definition for a set to be open in $\mathbb{R}^{n}$; but $\mathbb{R}^{n}$ is also open by definition and $\varnothing = \mathbb{R}^{n} \setminus \mathbb{R}^{n}$, so $\varnothing$ is closed in $\mathbb{R}^{n}$ by definition.

score 0 · Answer 4 · answered Oct 03 '15 at 08:31

In the definition of a topological space by open sets (there are others); the whole set and the empty set are always open by definition.

They are also closed.

Sets that are both open and closed, are in some texts called clopen sets.

Hence, for any topology on a given set, both the empty and whole set are clopen.

Is $\emptyset$ in $R^n$ an open set?

4 Answers4