If A⊂B,prove that $A^\circ⊂B^\circ$ and $cl (A) ⊂ cl(B)$

Question

For the open sets, $A ⊂ B, B^\circ ⊂ B, A^\circ ⊂ A$ We can say $A^\circ ⊂ B$ ..not sure what to do next.

For the closed sets, $A ⊂ cl(A) , B ⊂ cl (B)$ then $A ⊂ cl (B)$ and I'm not sure what do next?

Am I on the right track or I need to use the definitions for the proof?

The last step is to realize that $A^\circ$ is open and $B^\circ$ is the union of all open subsets of $B$. — Stefan Hamcke, Feb 17 '14 at 13:33
@Stefan, how come B^o is the union open subsets of B and it is a subset of B?? I don't get it. — Omni, Feb 17 '14 at 13:42
That's the definition of the interior. Or at least one possible definition. — Stefan Hamcke, Feb 17 '14 at 13:43
For the question about closure see: http://math.stackexchange.com/questions/121236/if-a-is-a-subset-of-b-then-the-closure-of-a-is-contained-in-the-closure-of-b — Martin Sleziak, Feb 17 '14 at 14:32

score 3 · Answer 1 · answered Feb 17 '14 at 13:42

Well ...

... or [do] I need to use the definitions for the proof?

yes. There might be some shortcut (which then would use the definition), but even then it's a good practice to use the definitions.

So, if you don't want to miss the practice, don't read the following spoiler.

In the first exercise it's near an immediate deduction from the very definitions:

\begin{align*} A^° = \bigcup_{U\in\mathcal{O}(X), U\subseteq A} U \end{align*}

therefore:

\begin{align*} x \in A^° \Rightarrow \exists U \in \mathcal{O}(X): U\subseteq A\wedge x \in U \end{align*}

and since $A \subseteq B$ it holds that $\exists U \in \mathcal{O}(X): U\subseteq A\subseteq B\wedge x \in U$, so $x \in B^°$.

Even in the second exercise it's just a small step -- but you may want to go by contradiction to keep it simple:

\begin{align*} x \notin \bigcap_{C\in\mathcal{C}(X), B \subseteq C} \Rightarrow \exists C \in \mathcal{C}(X) : B \subseteq C \wedge x \in C \end{align*}

and since $A\subseteq B$, also $A \subseteq C$ and therefore $x \notin cl(A)$.

score 1 · Answer 2 · answered Feb 17 '14 at 13:40

Just use the definitions. I assume you are talking about subsets of $\mathbb{R}^n$.

Let us prove $A^\circ \subset B^\circ$. Let $x\in A^\circ$. Then there is $\epsilon>0$ such that $B(x,\epsilon) \subset A\subset B$. Therefore $x\in B^\circ$.

Now the other part. First $A\subset B\subset \mathrm{cl}(B)$. The closure of $A$ is the intersection of all closed sets containing $A$. Since $\mathrm{cl}(B)$ is a closed set containing $A$, we get $\mathrm{cl}(A)\subset \mathrm{cl}(B)$.

score 1 · Answer 3 · answered Feb 17 '14 at 13:43

$ x \in A^\circ $ <=> there exists an open set $O_x$ containing $x$ in $A$. So we have $O_x \subset A \subset B$. Hence, by definition, $x \in B^{\circ}$.

For the other part we follows from the fact that : ${B^C}^\circ \subset {A^C}^\circ$.

Complement of these sets are closures of $B$ and $A$.

score 1 · Answer 4 · answered Feb 17 '14 at 13:44

1

Hint: You can use the following equivalent definitions: $A^o$ is the largest open set contained within $A$ and $cl(A)$ is the smallest closed set containing $A$. Have you proved these already / seen these yet?

answered Feb 17 '14 at 13:44

Gautam Shenoy

10,318

No, I haven't. Are those theorems? All I have learned so far is the definition of open sets and closed sets. – Omni Feb 17 '14 at 13:51
I am sure the other answers have solved your query. However I also urge you to show that the above statements are equivalent to whatever definitions you know of. Walter Rudin is a good source. The aforementioned definition is useful in your problem for it solves it instantly. Care to try first and see how? – Gautam Shenoy Feb 17 '14 at 13:56
The definition you mentioned is quite helpful and makes things easier. I already did it using the definition after I posted my question.I got similar proof to the ones posted here. I'll definitely check out the book though it is very hard to understand. I wish there was a simpler one. – Omni Feb 17 '14 at 14:03

If A⊂B,prove that $A^\circ⊂B^\circ$ and $cl (A) ⊂ cl(B)$

4 Answers4