# Proving that a metric space $X$ is open and closed subset of itself.

Let $$X$$ be a metric space. Let $$d$$ be its distance function.

Proving that $$X$$ is closed: Complement of $$X$$ relative to $$X=\emptyset$$, which is an open set (as all points of an open set are interior points and $$\emptyset$$ is empty and therefore vacuously $$\emptyset$$ is open) and hence $$X$$ is closed.

I am having difficulty in proving openness part.

Proving that $$X$$ is open: Let $$a\in X$$, then for any $$\epsilon \gt 0$$, neighborhood of $$a$$ is defined as $$N_\epsilon(a)=\{y\in X:d(a, y) \lt \epsilon\}$$
How do I claim that an $$\epsilon\gt 0$$ exist such that $$N_\epsilon (a) \subset X$$?

Edit: The definitions of Open and closed sets, which I have used here (let $$X$$ be a metric space) :

1. Open set: $$A\subseteq X$$ is an open set if for every $$a\in A$$, there exists a neighborhood $$N_r(a)$$ which lies completely in $$A$$ that is $$N_r(a) \subset A$$

2. Closed set: Complement of an open set in $$X$$

• By the definition of $N_{\epsilon}(a)$ you are restricting yourself to elements of $X$ – Jonathan Hole Aug 23 '20 at 22:35
• Choose your favourite $\epsilon>0$. It will satisfy the definition trivially. (eg $\epsilon = 1$ works, or $\epsilon = 10^{10000000}$ will also work, or $\epsilon = 10^{-100000000000}$ also works). All of these work, because $N_{\epsilon}(a)$ is by definition a subset of $X$, so there's nothing to be proven. – peek-a-boo Aug 23 '20 at 22:44
• You could do it by contradiction. Assume there exists an $\varepsilon>0$ such that $N_\varepsilon(a) \nsubseteq X$. Then there exists some $b \in N_\varepsilon(a)$ such that $b \notin X$. But $X$ is the whole space, which is a contradiction. Also, metric spaces endow a topology. By definition of a topology, the whole space must be open. – Ryan Aug 23 '20 at 22:53
• @PantelisSopasakis it doesn't matter. If $X=[0,1]$ with the usual metric, and $\epsilon = 10^{100}$, then $N_{\epsilon}(0.5) = [0,1]$ is the whole space. – peek-a-boo Aug 23 '20 at 22:55
• Clearly $N_\epsilon(x) \subset X$ for all $\epsilon,x$ and so $X$ is open. Since $X^C$ is vacuously open, the complement $X$ is closed. – copper.hat Aug 23 '20 at 23:05

Well, in a space $$X$$, the space contains all the points to be considered. There aren't any points not in the space $$X$$. So there aren't any points, limit points of $$X$$ or not, not in $$X$$ and every set whether an open neighborhood of a point $$x$$ or not is a subset of $$X$$.

If $$X$$ has any limit points or not, they are all in $$X$$ as there is nowhere else for them to be so $$X$$ is closed.

And for every point of $$x\in X$$ and every open neighborhood $$B_r(x) = \{!!!!\color{blue}{y\in X}!!!!| d(x,y) < r\} \subset \{\color{blue}{y\in X}\} = X$$. So every point $$x\in X$$ is an interior point of $$X$$. So $$X$$ is open.

.......

Oh, I see your definition of "closed" is not: $$A$$ is closed if all the limit points of $$A$$ are elements of $$A$$;

but is instead: $$A$$ is closed if it's complement is open.

Well, since $$X^c = \emptyset$$ we have to show $$\emptyset$$ is open. Your definition of open is: for every $$x \in \emptyset$$ ..... something. Well, since $$\emptyset$$ has no points and there are no $$x \in \emptyset$$ that is vacuously true.

So $$\emptyset=X^c$$ is open and $$X$$ is closed.

• " But it's difficult to visualize, I guess." It's intensely EASY to visualize. $X$ is the ENTIRE UNIVERSE there is !!!!!!!NOTHING!!!!!!!!! else. Every point that is in $X$ is ..... in $X$. Every subset of $X$.... is a subset of $X$. A neighborhood is defined as all the elements in $X$ that do something. Read that again, read that a thousand times. A neighboorhood is defined as all the element in $X$ that ..... all the elements in $X$ that do something... Well, if they are in $X$ then .... THEY ARE IN $X$!!!!!!!! – fleablood Aug 24 '20 at 17:14
• In your example, (0,0) is of course an interior point. Consider $N_{\epsilon}((0,0)) = \{x \in S| d(x,(0,0)) < \epsilon\}$. For insance if $\epsilon = 0.00001$ then $N_{\epsilon}((0,0)) = \{(0,0)\}$. Obviously $\{0,0\}\subset S$. But is it true for all $\epsilon$? Well, if $x \in N_{epsilon}((0,0))$ then $x \in S$ and $d(x,(0,0)) < \epsilon$. And .... !!!!$x \in S$!!!!!. So $N_{epsilon}((0,0))\subset S$. – fleablood Aug 24 '20 at 17:20
• To be an interior point there is no requirement that $N_{\epsilon} (a)$ contains any other point other than $a$. But if $x \in N_{\epsilon}$ then $x \in X$ because $N_{\epsilon}$ is BY DEFINITION a subset of $X$. You said what if $X$ were bounded or worse discreet... Okay, what if $X = \{5\}$. Claim: every point of $X$ is an interior point. Pf: If $x$ is a point of $X$ then $x=5$. And every neighborhood $N_{r}(5)=\{x\in\{5\}|d(x,5)<r\}=\{5\}$. And $\{5\}\subset \{5\}$. So $5$ is an interior point of $\{5\}$. – fleablood Aug 24 '20 at 17:35
• We can even have $X = \emptyset$. that if $x \in X$ then $N_{r}(x) = \{y\in \emptyset| d(x,y)<r\} = \emptyset \subset X$. Not a problem. – fleablood Aug 24 '20 at 17:37
• I like the way how fleablood emphasizes the important parts. – Hermis14 Jan 9 at 16:33

You give the definition yourself:

$$N_\varepsilon(a)=\{y\in X:d(a, y) \lt \epsilon\}$$

This set is by definition a subset of $$X$$ (because of the "$$y \in X$$" clause). So just pick $$\varepsilon =1$$ (any positive number will do) and note that

$$a \in N_1(a) \subseteq X$$ as required.

• I got confused between the definition of interior point and limit point so thought that $N_\epsilon (a) -\{a\}$ should be a subset of $X$. But now I have understood that this $"-\{a\}"$ is not required. Thanks a lot. – Koro Aug 24 '20 at 20:51
• Well, $N_{\epsilon}(a) -\{a\}\subset N_{\epsilon}(a)\subset X$. Note $N_{\epsilon}(a) -\{a\}$ could be empty. And empty sets are subsets of every set. – fleablood Aug 24 '20 at 21:16