Confusion regarding open and closed sets

Question

Definition: Let $(X, d)$ be a metric space. A set $A \subseteq X$ is open if $\forall x \in A \exists \varepsilon >0$ such that $B_{\varepsilon}(x) \subseteq A$. A set $O$ is open if and only if $X \setminus O$ is closed.

I am just confused regarding the last part. For example, let's let $X = \{0\} \cup \{1\} \cup \{2\} \cup (3, 7)$ and let $d$ be the usual distance metric in $\mathbb{R}$. Now say $O = \{0\}$, clearly this set is open since we can find an open ball around it (i.e., there exists a $\varepsilon>0$ such that $B_{\varepsilon}(0) = \{0\}$) which is contained in $\{0\}$. However, now consider the set $X \setminus \{0\} = \{1\} \cup \{2\} \cup (3, 7)$. But this set is not closed because $3$ is a limit point and $3$ is not included in the set. Can someone explain to me where my understanding has failed?

But I thought limit points do not have to be elements of the set? — user40333, Jul 26 '17 at 01:55
$3$ is not a limit point of $X\setminus{0}$ in $X$ because $3$ is not in $X$. A closed set in $X$ does not have to be a closed set in $\Bbb R$. — ajotatxe, Jul 26 '17 at 01:56

Aweygan · Accepted Answer · 2017-07-26T02:14:21.940

2

When we claim that a set is open or closed, it needs to be specified what the metric space in question is. If $(X,d)$ is a metric space and $A\subset X$, then $(A,d)$ is also a metric space. However, a subset of $A$ may be open or closed in $A$, while failing to be open or closed in $(X,d)$.

When we say that $O=\{0\}$ is open, it is as a subset of $X$, not as a subset of $\mathbb R$ (which is clearly false). So we have $X\setminus O$ is closed in $X$, but again, not in $\mathbb R$.

edited Jul 26 '17 at 02:14

answered Jul 26 '17 at 02:08

Aweygan

23,232

Oh, I think I understand now. For any point $x_L$ to be a limit point of some subset of $X$, first we must have $x_L \in X$. So for example, if $X = \mathbb{R}$ here, then clearly the set $A = {1} \cup {2} \cup (3,7)$ is not closed (and ${0}$ would not be open either). But if we only considered the metric space $X$ as given in my OP, then in fact ${1} \cup {2} \cup (3,7)$ is closed (in $X$) because $3$ can't be a limit point of it, since $3$ is not an element of our metric space. Is this right? – user40333 Jul 26 '17 at 02:14
Yep, that's corect. – Aweygan Jul 26 '17 at 02:15

score 0 · Answer 2 · 2017-07-26T02:22:58.200

0

Edit:

I misread the question.

$X\backslash\{0\}$ is closed, and $3$ is not a limit point of it, because $3$ doesn't even lie in the metric space $X$.

edited Jul 26 '17 at 02:22

answered Jul 26 '17 at 02:02

Can't I just set $\varepsilon = d(0,1)$, so then $B_{\varepsilon}(0) = {0}$? – user40333 Jul 26 '17 at 02:08
Yes, that will be correct. – Jul 26 '17 at 02:23

Confusion regarding open and closed sets

2 Answers2