# Is the whole set $\mathbb R$ open?

Is the whole set $\mathbb R$ open or closed? A lot of answers from the following link said that the whole set is closed.

Why is empty set an open set?

However varies notes said that $\mathbb R$ is open, for example:

https://www.math.cornell.edu/~hatcher/Top/TopNotes.pdf

• Open does not mean not closed. – Improve Jun 29 '17 at 2:58
• It is both... some say "clopen" the empty set is the another classic example of a set that is both open and closed. – Doug M Jun 29 '17 at 3:24
• "Is the whole set R open or closed?" both. "Said that the whole set is closed". That is true. "However varies notes said that R is open". That is also true. R is open. And R is closed. – fleablood Jun 29 '17 at 4:00

It's open and closed by definition.

In order to refer to open and closed sets a topology must be made explicit. Given a set $X$ and a collection $\mathcal F$ of subsets of $X$, $\mathcal F$ is a topology on $X$ only if $X, \emptyset \in \mathcal F$. Thus X is open by definition, and as closed sets are defined as sets whose compliments are open we have $X^c=\emptyset$ and we have $X$ is closed.

In the context of a topological space $\mathbb R$ with collection $\mathcal F$, $\mathbb R$ must be open and closed. However if we consider $\mathbb R \times \mathbb R$ with the usual topology, $\mathbb R \simeq \{0\} \times \mathbb R$ is closed but not open. It is important to state context.

Indeed the observation by Michael Hardy does really make sense. Additionally the question is not well posed because it does not state which topology is to be considered. Let us assume that euclidean topology is meant. So the answer by Michael Rozenberg uses axiomatization via open or closed sets.

Using axiomatization via neighborhoods (read balls in this case, due to the euclidean topology of $\mathbb R$)

1. $\mathbb R$ is open because any of its points have at least one neighborhood (in fact all) included in it;

2. $\mathbb R$ is closed because any of its points have every neighborhood having non-empty intersection with $\mathbb R$ (equivalently punctured neighborhood instead of neighborhood).

Equivalently:

1. $\mathbb R$ is open because all its points are interior points of itself

2. $\mathbb R$ is closed because all its points are adherent points of itself (equivalently limit points instead of adherent points)

Using axiomatization via Moore-Smith net convergence (read sequence convergence in this case, due to the euclidean topology of $\mathbb R$),

1. $\mathbb R$ is closed because every point to which at least one net of its points converges belongs to it

2. $\mathbb R$ is open because every point to which every convergent net converges has a non-empty intersection with it. (Or equivalently there are no nets of points of its complement (the empty set) converging to any of its points (in fact there are no nets of points of its complement))

The list can go on and on.

By the definition of a topology on a set, both the empty set and the entire set (which I'm assuming you're taking as $\mathbb R$) are open sets. Since the complement of an open set is a closed set, and $\mathbb R$ is the complement of the empty set, it is also closed.

Notice that "open" and "closed" are not mutually exclusive.

$A$ is open iff $A^c$ is closed.

So $\emptyset$ and $\Omega$ are both open and closed (clopen)

For general topology, it is guaranteed via definition of topology, i.e. $\emptyset, \Omega \in \mathscr T$.

By one commonplace definition of "open set", a set $A\subseteq\mathbb R$ is open if for every point $x\in A$ there is some open interval containing $x$ that is a subset of $A$. That clearly is true if $A=\mathbb R,$ so $\mathbb R$ is open.

By one commonplace definition of "closed set", a set $A\subseteq \mathbb R$ is closed if every limit point of $A$ is a member of $A$. That clearly is true if $A=\mathbb R$, so $\mathbb R$ is closed.

Only two subsets of $\mathbb R$ are both open and closed: $\mathbb R$ and $\varnothing.$