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

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

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

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

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

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

(2) $\overline{the intersection of A_alpha}$= $\bigcap$ $\overline{A}$_ $\alpha$

My Work: $\overline{the intersection of A_alpha}$ $\subset$ $\bigcup$A_ $\overline{alpha}$. I only need to show $\subset$ 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) $\overline{A-B}$= $\overline{A}$- $\overline{B}$.

My Work: $\overline{A-B}$ $\cup$ $\overline{B}$= $\overline{A-B UNION B}$= $\overline{A UNION B}$ $\supset$ $\overline{A}$ $\Rightarrow$ $\overline{A-B}$ $\supset$ $\overline{A}$- $\overline{B}$. ...A:=(0,2) B:=(1,2) $\overline{A-B}$:=[0,1] $\overline{A}$- $\overline{B}$:=[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 $x \in \overline{A\cap B}$ then $(i) x \in A\cap B \vee (ii) \forall r > 0: N_r(x)\cap(A\cap B)\setminus\{x\} \ne \emptyset$.

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

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

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

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

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

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

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

4. For (3) I belive you are wrong. Since $\overline{A - B} \supset \overline{A} - \overline{B}$ 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 $\overline{A - B} \supset \overline{A} - \overline{B}$ 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.. $\overline{R-Q}$=R $\neq$ $\overline{R}$- $\overline{Q}$=(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.