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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 $X$

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    $\begingroup$ By the definition of $N_{\epsilon}(a)$ you are restricting yourself to elements of $X$ $\endgroup$ – Jonathan Hole Aug 23 '20 at 22:35
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    $\begingroup$ 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. $\endgroup$ – peek-a-boo Aug 23 '20 at 22:44
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    $\begingroup$ 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. $\endgroup$ – Ryan Aug 23 '20 at 22:53
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    $\begingroup$ @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. $\endgroup$ – peek-a-boo Aug 23 '20 at 22:55
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    $\begingroup$ 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. $\endgroup$ – copper.hat Aug 23 '20 at 23:05
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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.

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    $\begingroup$ " 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 $X$!!!!!!!! $\endgroup$ – fleablood Aug 24 '20 at 17:14
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    $\begingroup$ 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 $N_{epsilon}((0,0))\subset S$. $\endgroup$ – fleablood Aug 24 '20 at 17:20
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    $\begingroup$ 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 $\{5\}$. $\endgroup$ – fleablood Aug 24 '20 at 17:35
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    $\begingroup$ 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. $\endgroup$ – fleablood Aug 24 '20 at 17:37
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    $\begingroup$ I like the way how fleablood emphasizes the important parts. $\endgroup$ – Hermis14 Jan 9 at 16:33
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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.

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  • $\begingroup$ 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. $\endgroup$ – Koro Aug 24 '20 at 20:51
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    $\begingroup$ 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. $\endgroup$ – fleablood Aug 24 '20 at 21:16

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