If A is a subset of a topological space, then $Bd(A) \subseteq Cl(A)$. Prove using elements/ sets.

If A is a subset of a topological space, then $Bd(A) \subseteq Cl(A)$. Prove.

I know this statement is true. I am now trying to prove it. I am in a basic topology class and to do a lot of set proofs we start by letting an element be included in one side and show it is in the other.

I was given the hint to use this fact $Cl(A)= A \cup Bd(A)$. However I have to prove that statement first. For this statement I started with Let $x \in Cl(A)$ then by defintion of closure $x \in A$ since the closure is the smallest closed set containing $A$ and therefore $x \in A \cup Bd(A)$. Now let $x \in A \cup Bd(A)$. How do I show $x \in$ the $Cl(A)$ and then use that prove my first statement?

• Did you take a Set Theory class before diving into topology? – user230734 Apr 28 '15 at 21:15
• How do you define $Bd(A)$ (I guess it is the border of $A$)? – Berci Apr 28 '15 at 21:16
• What is your definition of boundary? I usually see the boundary defined as the intersection of the closure of $A$ and the closure of the complement of $A$ or something equivalent. – jgon Apr 28 '15 at 21:16
• Another (equivalent) definition of the boundary is $\partial A=\overline{A}\setminus A^\circ$ (where $\partial A$ is the boundary, $\overline{A}$ is the closure, and $A^\circ$ the interior). With this definition the result in question is trivial. – Hayden Apr 28 '15 at 21:42
• Our introduction to topology book chapter 1 was set theory while chapter 2 was a basic introduction to topology – user219081 Apr 28 '15 at 22:22

I am not sure I understand the hint. If you are allowed to assume it, the question is trivial, but proving it is equivalent to proving the problem...

Let $x\in Bd(A)$. Then by definition, every open set containing $x$ intersects both $A$ and $X−A$.

If $x\notin Cl(A)$, then there exist on open set, namely $U = X-Cl(A)$, containing $x$ such that $U\cap A = \emptyset$.

We can see this since $A\subset Cl(A)$.

But this contradicts the definition of $x\in Bd(A)$, so we get that $x\in Cl(A)$.

Thus, $Bd(A)\subset Cl(A)$.

• I think we were supposed to prove the hint in order to use it to prove the statement. – user219081 Apr 28 '15 at 22:37
• Right, but your approach of showing $A\cup Bd(A) \subset Cl(A)$ forces you to show $Bd(A)\subset Cl(A)$. So proving your hint is actually more work... – WSL Apr 28 '15 at 22:39
• I can see that now... – user219081 Apr 28 '15 at 22:48
• Does this argument make sense? – WSL Apr 28 '15 at 22:49
• Yes but I have never proved that $A \subset Cl(A)$ so I think I would either need to prove that or use something else. – user219081 Apr 28 '15 at 23:17