3
$\begingroup$

Let $A$ be a subset of a topological space. Prove that $Cl(A) = Int(A) \cup Bd(A)$

Here are my defintions:

Closure: Let $(X,\mathfrak T)$ be a topological space and let $ A \subseteq X$ . The closure of $A$ is $Cl(A) = \bigcap \{U \subseteq X: U$ is a closed set and $A \subseteq U\}$ Based on this I know $A \subseteq Cl(A)$

Interior:Let $(X, \mathfrak T)$ be a topological space and let $A \subset X$ is the set of all points $x \in X$ for which there exists an open set $U$ such that $x \in U \subseteq A$.

My definition of boundary is: Let $(X,\mathfrak T)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is in the boundary of A if every open set containing $x$ intersects both $A$ and $X−A$.

My proofs start by picking an element to be in each side then showing it must be in the other side. I have tried to start that here.

Let $x \in Cl(A)$ then

Let $x \in Int(A) \cup Bd(A)$. Then $x\in Int(A)$ or $x\in Bd(A)$ If $x \in Int(A)$ then

$\endgroup$
4
  • 1
    $\begingroup$ Can you use the definition $\text{Bd}(A) = \overline{A}\setminus \text{Int}(A)$? If so, this proof is a one-liner. $\endgroup$
    – graydad
    Commented May 15, 2015 at 19:38
  • $\begingroup$ so I did that and I got to here: $Cl(A) = Int(A) \cup ((Cl(A) - Int(A))$ Now what? $\endgroup$
    – user219081
    Commented May 15, 2015 at 19:48
  • $\begingroup$ can I just conclude then $Cl(A) = Cl(A)$? $\endgroup$
    – user219081
    Commented May 15, 2015 at 19:48
  • 1
    $\begingroup$ Yup! It really is that easy. By set algebra, you have equality. It's kind of like saying $X = Y+(X-Y)$. I will write up an answer though using the definitions you have above, if you'd like. It is still good to be comfortable with multiple definitions :) $\endgroup$
    – graydad
    Commented May 15, 2015 at 19:49

3 Answers 3

1
$\begingroup$

You can do it directly, after showing that $x\in\operatorname{Cl}(A)$ if and only if, for every neighborhood $U$ of $x$, $U\cap A\ne\emptyset$.

Suppose $x\in\operatorname{Cl}(A)$. If there exists a neighborhood $U$ of $x$ such that $U\subseteq A$, then $x\in\operatorname{Int}(A)$. Otherwise, no neighborhood of $x$ is contained in $A$, so every neighborhood of $x$ interesects $X\setminus A$, which means $x\in\operatorname{Bd}(A)$.

Thus we have proved that $\operatorname{Cl}(A)\subseteq\operatorname{Int}(A)\cup\operatorname{Bd}(A)$.

Conversely, it is clear that $\operatorname{Int}(A)\subseteq A\subseteq\operatorname{Cl}(A)$. Also $\operatorname{Bd}(A)\subseteq\operatorname{Cl}(A)$ is clear, because every neighborhood of a point $x\in\operatorname{Bd}(A)$ intersects $A$ (and $X\setminus A$).

$\endgroup$
0
$\begingroup$

The closure of $A$ is the smallest closed subset of $X$ containing $A$. The interior of $A$ is the largest open subset contained in $A$. So we have that $Int(A) \subseteq A \subset Cl(A)$. Let $x \in Bd(A)$. Suppose $ x \notin Cl(A)$. $X \setminus Cl(A)$ is open, and $x \in X \setminus Cl(A)$, so by your definition of boundary, $(X \setminus Cl(A)) \cap A \neq \emptyset$, which contradicts that $ A \subseteq Cl(A)$. So $Cl(A) \supseteq Int(A) \cup Bd(A)$.

For the converse we show that $D=Int(A) \cup Bd(A)$ is closed and contains $A$. Suppose $a \in A$ but $ a \notin D$. In particular $a \notin Bd(A)$ so there is an open set $U_{a} \subseteq A$ which does not intsersect $X \setminus A$. But then $U_{a}$ is an open set contained in $A$, so $U_{a} \subseteq Int(A)$ a contradiction. So $A \subseteq D$. For each $x \in X \setminus D$, there is an open set $U_{x} \subseteq X\setminus D \subseteq X \setminus A$ containing $x$, because otherwise $x \in Bd(A)$. Taking the union of these open subsets for each $x \in X\setminus D$ gives us that $X \setminus D$ is open, so $D$ is closed. $\square$

$\endgroup$
1
  • 1
    $\begingroup$ Closure is smallest, not largest. $\endgroup$
    – Adayah
    Commented May 15, 2015 at 20:14
0
$\begingroup$

Here's how I would go about it.

Let $x \in \text{Bd}(A) \cup \text{Int}(A)$. If $x \in \text{Bd}(A)$ then $x$ is a limit point because all boundary points are limit points. Then $x \in \overline{A}$. Else if $x \in \text{Int}(A)$ it is also clear that $\text{Int}(A) \subseteq \overline{A}$ so $x \in \overline{A}$. Hence $\text{Bd}(A) \cup \text{Int}(A) \subseteq \overline{A}$.

Now let $x \in \overline{A}$. We know that either $x \in \text{Int}(A)$ or $x \notin \text{Int}(A)$. We have interest only in the latter case. Suppose toward contradiction that $x$ is not a boundary point of $A$. Then there exists an open set $U$ containing $x$ such that either $U\cap A = \emptyset$ or $U\cap(X\setminus A) = \emptyset$. We can dismiss the first case, because that would imply $x$ is not a limit point of $A$ and hence $x\notin A$, so it must be that $U\cap(X\setminus A) = \emptyset$. Clearly this means $x \notin \emptyset =U\cap(X\setminus A)$ so $x$ is in the complement, $U^c \cup A$ (you can prove this is the complement with Demorgan's law and a little set algebra). We know $x \notin U^c$ since $x \in U$. Then $x \in U^c \cup A$ means $x \in A$. Further, the complement of the emptyset is $X$, so we know $X = U^c \cup A$. Since $X$ is open and $U^c$ is closed, $X\setminus U^c = A$ is open, meaning $A = \text{Int}(A)$. Hence, $x \in \text{Int}(A)$, a contradiction. We conclude that $\overline{A} \subseteq \text{Bd}(A) \cup \text{Int}(A)$.

$\endgroup$
3
  • $\begingroup$ I think I might have to go about it this way. I get a little confused by what we are allowed to use as definitions and what we are not allowed to use. Did you see my other question about the half open interval topology today? Can you give me any insight there? I feel like you are a gigantic help! $\endgroup$
    – user219081
    Commented May 15, 2015 at 20:14
  • 1
    $\begingroup$ @AlyssaWallace I mean, I personally think any definition should be fair game! After all, definitions are equivalent. You can't have a point be a boundary point under one definition and not a boundary point under another definition (for example). The only time I'd stick to one definition is if your professor requires it or something. Sometimes one definition makes a proof much smoother than another. As for your other question today, I did see! I have not encountered that topology and need some more time to mull it over. I'll post an answer if I figure it out though. $\endgroup$
    – graydad
    Commented May 15, 2015 at 20:23
  • 1
    $\begingroup$ @AlyssaWallace proof fixed. $\endgroup$
    – graydad
    Commented May 15, 2015 at 20:39

You must log in to answer this question.