# Thread: Determine whether the equations hold (equality, inclusion etc.)

1. ## Determine whether the equations hold (equality, inclusion etc.)

Let A, B, and A_$\displaystyle \alpha$ denote subsets of a space X. Determine whether the following equations hold, if an equality fails, determine whether one of the inclusions $\displaystyle \subset$ or $\displaystyle \supset$ holds.

(1) $\displaystyle \overline{AintersectB}$=$\displaystyle \overline{A}$$\displaystyle \cap$$\displaystyle \overline{B}$

My Work: A$\displaystyle \cap$B$\displaystyle \subset$A,B. Thus A$\displaystyle \cap$B$\displaystyle \subset$$\displaystyle \overline{A},\displaystyle \overline{B} and therefore A\displaystyle \capB\displaystyle \subset$$\displaystyle \overline{A}$$\displaystyle \cap$$\displaystyle \overline{B}$. Thus $\displaystyle \overline{AintersectB}$$\displaystyle \subset$$\displaystyle \overline{A}$$\displaystyle \cap$$\displaystyle \overline{B}$

Is this enough? Or do I need further information explaining it, that the part I always lack in my proofs.

(2)$\displaystyle \overline{the intersection of A_alpha}$=$\displaystyle \bigcap$$\displaystyle \overline{A}_\displaystyle \alpha My Work:\displaystyle \overline{the intersection of A_alpha}$$\displaystyle \subset$$\displaystyle \bigcupA_\displaystyle \overline{alpha}. I only need to show \displaystyle \subset right because I only need to show what inclusion works if equality fails? i'm having the most trouble with this one because i'm very unsure about it. (3) \displaystyle \overline{A-B}=\displaystyle \overline{A}-\displaystyle \overline{B}. My Work: \displaystyle \overline{A-B}$$\displaystyle \cup$$\displaystyle \overline{B}=\displaystyle \overline{A-B UNION B}=\displaystyle \overline{A UNION B}$$\displaystyle \supset$$\displaystyle \overline{A} \displaystyle \Rightarrow$$\displaystyle \overline{A-B}$$\displaystyle \supset$$\displaystyle \overline{A}$-$\displaystyle \overline{B}$. ...A:=(0,2) B:=(1,2) $\displaystyle \overline{A-B}$:=[0,1] $\displaystyle \overline{A}$-$\displaystyle \overline{B}$:=[0,1). Is this correct? and/or enough? or again do I need more explanation.

***I am sorry in advanced for my bad latex skills I'm learning and tried very hard this time to make it as clear as I could with what i know.***

2. What you did is not really a proof, in my opinion. Make sure to use the definitions of the particular items you are using. For (1):

If $\displaystyle x \in \overline{A\cap B}$ then $\displaystyle (i) x \in A\cap B \vee (ii) \forall r > 0: N_r(x)\cap(A\cap B)\setminus\{x\} \ne \emptyset$.

If (i): $\displaystyle \overline{A}\cap\overline{B} \supset A\cap B \ni x$.
If (ii): $\displaystyle N_r(x)\cap A\setminus\{x\} \ne \emptyset \wedge N_r(x)\cap B\setminus\{x\} \ne \emptyset \Rightarrow x\in \overline{A} \wedge x \in \overline{B} \Rightarrow x \in \overline{A}\cap\overline{B}$

Thus: $\displaystyle \overline{A\cap B} \subset \overline{A}\cap \overline{B}$

If $\displaystyle x \in \overline{A}\cap\overline{B}$ then $\displaystyle (i) x \in A\cap B \vee (ii) \forall r > 0: N_r(x)\cap A\setminus\{x\} \ne \emptyset \wedge N_r(x)\cap B\setminus\{x\}$.

If (i): $\displaystyle \overline{A\cap B} \supset A\cap B \ni x$.
If (ii): You must ask yourself, can there be limit points of both $\displaystyle A$ and $\displaystyle B$ that are no limit points of $\displaystyle A\cap B$. The answer is no again since: $\displaystyle \emptyset \ne N_r(x)\cap A\setminus\{x\} \cap N_r(x)\cap B\setminus\{x\} = N_r(x)\cap (A\cap B\setminus\{x\}) \Rightarrow x \in \overline{A\cap B}$

Thus: $\displaystyle \overline{A\cap B} \supset \overline{A}\cap \overline{B}$ and therefore $\displaystyle \overline{A\cap B} = \overline{A}\cap \overline{B}$

3. If x is a limit point of $\displaystyle \cap A_{\alpha}$ then this limit point must have neighbourhoods that allways contain points of this intersection. Since we look at an intersection, all $\displaystyle A_{\alpha}$ must contain the same elements and therefore this limit point is a limit point of any $\displaystyle A_{\alpha}$. It follows that $\displaystyle \overline{\cap A_{\alpha}} \subset \cap \overline{A_{\alpha}}$.

On the other side. Imagine $\displaystyle A_{\alpha} = (0,\frac{1}{\alpha})$ with $\displaystyle \alpha \in \mathbb{N}$. Then $\displaystyle \cap A_{\alpha} = \emptyset$. But the closure of each $\displaystyle A_{\alpha}$ contains the element zero. Thus the intersection of all closures will contain 0. Thus the intersection of all closures is not empty. Thus $\displaystyle \overline{\cap A_{\alpha}} \not\supset \cap \overline{A_{\alpha}}$ in the general case.

4. For (3) I belive you are wrong. Since $\displaystyle \overline{A - B} \supset \overline{A} - \overline{B}$ but not necessarily the other way round. Think of an counter example. The proof for the former only contains steps I showed you before so try to solve this yourself first.

5. Originally Posted by raphw
For (3) I belive you are wrong. Since $\displaystyle \overline{A - B} \supset \overline{A} - \overline{B}$ but not necessarily the other way round. Think of an counter example. The proof for the former only contains steps I showed you before so try to solve this yourself first.
hmm..$\displaystyle \overline{R-Q}$=R$\displaystyle \neq$$\displaystyle \overline{R}$-$\displaystyle \overline{Q}$=(Empty Set)?

R and Q as in the reals and the rationals

6. Yep. Good example! You can still prove that the subset constraint I posted above holds, however.