# Prove that for A $\subseteq$ B int(A) $\subseteq$ int(B) and cl(A) $\subseteq$ cl(B).

I have to prove that for A$$\subseteq$$B, int(A) $$\subseteq$$ int(B)and cl(A) $$\subseteq$$ cl(B). I hope someone here can help me out, and I apologize for any obvious mistakes. So here is my approach.

By definition, the interior of B int(B) is the largest open set contained in B. It is the union of all open sets in B: $$int(A)=_{W\subseteq A:\ W\;is\;closed}W$$. Thus if int(B) $$\subseteq$$ B and A $$\subseteq$$ B, then int(A) $$\subseteq$$ int(B).

By definition, cl(B), is the smallest closed set containing B. It is the intersection of all closed sets containing B. Thus it must hold that cl(A) $$\subseteq$$ cl(B), as A $$\subseteq$$ cl(A) and B $$\subseteq$$ cl(B), with A $$\subseteq$$ B.

• I think you have some typos in your first proof. The second one: you know that $A \subseteq \mbox{cl}A$ and $B\subseteq \mbox{cl}B$ but you don't know if $\mbox{cl}A \subseteq B$ to conclude $\mbox{cl}A \subseteq \mbox{cl}B$, right? Sep 16, 2020 at 0:17
• $W$ are open sets with $W\subseteq A$. Sep 16, 2020 at 0:17
• For the second portion of your question, see: math.stackexchange.com/questions/121236/… Sep 16, 2020 at 0:24
• $\text{Int}(A)$ is an open set and we have $\text{Int}(A) \subseteq A \subseteq B$; $\text{Cl}(B)$ is a closed set and we have $A \subseteq B \subseteq \text{Cl}(B)$ Sep 16, 2020 at 0:24

Hints:

(a) If $$U$$ is an open set such that $$U \subset A$$ and $$A\subseteq B$$ then $$U\subseteq B$$. By definition, $$\mbox{int}A$$ is the union of all such open sets. Now, it is an open set, which is contained in $$B$$, right? What can you conclude?

(b) Analogously, if $$F$$ is a closed set containing $$B$$, then it must contain $$A$$. Thus, $$\mbox{cl}B$$ is the intersection of all such closed sets. This a closed set and it contains $$B$$. What can you conclude?

Hint for the 1st. If $$x\in{\rm int}A$$ then exists an open $$W$$ with $$a\in W\subseteq A$$, but $$A\subseteq B$$ so $$x\in W\subseteq B$$, therefore $$x\in{\rm int}B$$. This settles $${\rm int}A\subseteq{\rm int}B$$.

$$\operatorname{int}(A)$$ is the maximal open set subset of $$A$$, so if $$A \subseteq B$$ it's in particular some open subset of $$B$$ too and hence a subset of the maximal open subset of $$N$$ which is by definition $$\operatorname{int}(B)$$. So $$\operatorname{int}(A) \subseteq \operatorname{int}(B)$$ follows.

Dually, $$\overline{B}$$ is the minimal closed superset of $$B$$, and hence also some closed superset of $$A$$ when $$A \subseteq B$$. As $$\overline{A}$$ is the minimal superset of $$A$$, again $$\overline{A} \subseteq \overline{B}$$ follows by minimality.

• Quick question: Is it true that $\overline{A} \subseteq \overline{B} \implies$ $A \subseteq B$? Nov 2, 2020 at 22:44
• @TaylorRendon no, consider the rationals and the irrationals. Nov 3, 2020 at 5:30

i recommend a pointwise approach. In other words, let $$a \in int(A)$$. Then by definition of being in the interior, we have that $$\exists U$$ open, such that, $$a \in U \subseteq A$$. Since $$A \subseteq B$$ it follows that $$a \in U \subseteq B$$, i.e. $$a \in int(B)$$.

Note that for the second question: $$cl(B)$$ is a closed set and $$A \subset B \subset cl(B)$$. So by definition and from what we are given we know that $$A \subset cl(A) \subseteq B \subset cl(B)$$.

So, $$cl(A) \subseteq cl(B)$$.

For the first question, it might be easier to approach in the following way to show that given $$A \subseteq B$$, $$int(A) \subseteq int(B)$$:

Let $$x \in int(A)$$. Then by definition we know that for some $$\epsilon > 0$$ there exists an open ball $$B_{\epsilon}(x) \subseteq A$$. Since we are given that $$A \subseteq B$$ and now know that $$B_{\epsilon}(x) \subseteq A$$ $$\implies$$ $$B_{\epsilon}(x) \subseteq B$$ so $$x \in int(B)$$. So $$int(A) \subseteq int(B)$$.