# The whole space is an open set in a metric space

Let $(X,d)$ be a metric space. Then both $X$ and $\emptyset$ are open sets.

I went over the proofs and it seems to be trivial, and I get it for $\emptyset$, but why must the whole space be open? Can't $X$ be a closed ball?

• Being closed or open is not an intrinsic property. The topology of a space does not know if the space itself is embedded somewhere else, nor what being open or closed in that topology means.
– user228113
Commented May 1, 2017 at 22:22
• $X$ is always both open and closed in any topology.
– JMJ
Commented May 1, 2017 at 22:23
• In a topological space, $X$ and $\emptyset$ are both open and closed. Sets can be both open and closed (the properties are not exclusive). Such sets are called clopen. Commented May 1, 2017 at 22:23
• A metric space is a type of topological space. Commented May 1, 2017 at 22:27
• What is your definition of "open"? Use that. Also, there is nothing wrong with something being simultaneously open and closed. Commented May 1, 2017 at 22:27

If $(X,d)$ is a metric space, then as far as closed and open sets in $X$ go, $X$ is the universe. There is nothing outside of $X$. There is nothing which $X$ is a part of. There is no intrinsic notion of a "boundary" of an arbitrary metric space.

Here is why $X$ is an open subset of $X$. Let $x \in X$. To show $X$ is open in $X$, we need to show there is at least one $\epsilon > 0$ such that the ball $B(x,\epsilon)$ with center $x$ and radius $\epsilon$ is contained in $X$. Note that by definition,

$$B(x,\epsilon) = \{ y \in X : d(x,y) < \epsilon \}$$

so $B(x,\epsilon)$ is always a subset of $X$, for any choice of $\epsilon$ whatsoever. This is why $X$ is open in $X$.

Example: Let $X = \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \}$ be the closed disc in $\mathbb{R}^2$. Give $X$ the Euclidean metric. If you consider $X$ as a subset of the metric space $\mathbb{R}^2$ (with the same metric), it is obviously not open. But if you consider $X$ as a subset of the metric space $X$, it is open.

If you are confused, think about the fact that open balls (and more generally open sets) in $X$ are in general not open when you consider them as subsets of $\mathbb{R}^2$. take the point $(1,0)$ in $X$ and consider the ball in $X$ of radius $\frac{1}{2}$ and center $(1,0)$. Draw it. By definition, it is an open set in the metric space $X$. But it's not going to be an open circle. And it will not be open when considered as a subset of $\mathbb{R}^2$.

• "By definition, it is an open set in the metric space X" that is because the ball contains every point of $X$ even if the ball is "cutted"?
– gbox
Commented May 1, 2017 at 22:46
• What? It doesn't contain every point of $X$.
– D_S
Commented May 1, 2017 at 22:47
• if X does not "live" in $\mathbb{R}^2$
– gbox
Commented May 1, 2017 at 22:48
• I don't understand what you're saying.
– D_S
Commented May 1, 2017 at 22:49
• I drew a picture of the ball of center $(1,0)$ and radius $\frac{1}{2}$ in $X$.
– D_S
Commented May 1, 2017 at 22:58

If $x \in X$ is any point and $\epsilon > 0$ any positive real number, then the open ball of radius $\epsilon$ centered at $x$, $\{ y\in X \mid d(x,y) < \epsilon)\}$, is by definition contained in $X$. Note that there is nothing wrong with a subset being simultaneously closed and open in a space (despite the terminology).

For example, consider the boundary point $(1,0)$ of the closed disk $$D = \{(x,y) \in \mathbb{R}^2\mid x^2 + y^2 \leq 1\}.$$ By definition, the open ball with respect to the metric of the closed disk centered around $(1,0)$ of radius $\epsilon > 0$ is the set of points in $D$ that are at distance $< \epsilon$ from $(1,0)$. (This is the red subset of the disk in the picture. The line has length $\epsilon$.)

For instance, if $\epsilon = 1000$, then this ball is the whole disk $D$. But no matter the choice of $\epsilon$, the ball will always be a subset of the disk $D$.

On the other hand, if we consider open balls with respect to the metric of the plane, then any open ball around $(1,0)$ won't be contained inside $D$.

• But what about the point in the boundary of the space? there will be no ball that will satisfy the requirement
– gbox
Commented May 1, 2017 at 22:34
• @gbox Yes, any ball of any radius centered around that point will do. Of course a closed ball is not going to be open in the plane, but it is always going to be open in itself. Note that the "open ball" in this case is the set of points inside the closed ball that are distance less than $\epsilon$ from the chosen boundary point; the metric knows nothing about points outside of the closed ball. Commented May 1, 2017 at 22:37
• "But what about the point in the boundary of the space? there will be no ball that will satisfy the requirement" ALL balls satisfy the requirement! A "ball" around y ={x in X| d (x,y) < e} $\subset$ X. Even if the point is on the boundary of the space, then then ball includes boundary of the space. Consider 1 in [0,1]. A ball around 1 is all points close to 1. That is (1-e,1]. It contains all the points that are close to the left. And it contains all points that are close to the right. It's just that there are no points to the right. A ball doesn't have to be "round". Commented May 1, 2017 at 23:08
• @gbox If $X$ has a boundary, that's not a problem. Looking at the example unit disc in the answer here, while we draw it as a subset of the plane, there is nothing "outside" $X$, whatever that would mean. $X$ is the collection of all points that exist as far as this problem is concerned; there​ are no other points that an open ball could contain. Commented May 1, 2017 at 23:15

A set that contains everything must be open as every neighborhood of every point must be a subset of the space. So every point is an interior point.

Take for example the metric space $[0,1]$ (this is a "closed ball" obviously). Then $N_e (1)=\{x\in [0,1]| d (x,1)< e\}=(1-e] \subset [0,1]$. So 1 is an nterior point.

I have also seen a lot of "trivial" answers, which somehow look incomplete. I like to prove it with a contradiction. The theorem states:

$$X \subseteq X$$ is open because $$\forall x \in X : \exists \epsilon >0 : B_\epsilon (x) \subseteq X$$.

Suppose $$X \subseteq X$$ is not open. Then we logically negate the "open set" condition:

$$\exists x \in X : \forall \epsilon > 0 : B_\epsilon (x) \cap (X \setminus X) \ne \emptyset$$

In words: we expect points on the boundary span non-pointlike balls which should intersect with whatever is beyond the set X. Now, X contains all elements. In $$(X, d)$$ X is our universe. Any other set within can be equal or smaller than X. Any other set beyond is empty.

So, let's find that intersection. Firstly, $$X \setminus X = \emptyset$$, since X is the universe. Now comes the funny part. Any intersection with empty set is empty, since empty set does not possess any elements. And, if intersection is empty, then those sets do not overlap. And this is a contradiction.

Probably, the confusion stems from the symmetry of the boundary of the ball. The "open ball" set does not have to be filled homogeneously as a ball. The "open ball" is a ball only to the extent the underlying set is providing non-empty elements to fill that ball.