Vtyjkyuk

Let $F$ be collection of set in $R^n$ and let $S=cup_Ain FA$ and $T=cap_Ain FA$.

Then prove or disprove following facts.

1) If x is limit point of T,then x is limit point of each of $Ain F$


2) If x is limit point of S,then x is limit point of atleast one of $Ain F$.

My Attempt:
1) x is limit point of $cap_Ain FA$ then for $forall epsilon>0 exists yin $$cap_Ain FA$ i.e $y in $every$ A$ Hence x is limit point of every A.

2)x is limit point of $cup_Ain FA$ then for $forall epsilon>0 exists yin $ A for some $A in F$ Hence done.

Actually both statement looking like trivial.

I just wanted to confirm are they true.Or there exists some counterexample.

ANy Help will be appreciated

If the statements are looking trivial, then you are not looking carefully!

What does it mean to be a limit point of a set? $x$ is a limit point of a set $B$ if for all $epsilon > 0$ there exists $y in B$ such that $d(y,x) < epsilon$. (Sometimes, $x neq y$ is also insisted).

The intersection is done correctly : if $x$ is a limit point of $cap_A in F A$, then for all $epsilon > 0$ there is $y in cap_A in F A$, such that $d(y,x) < epsilon$. Now, note that $y in A$ for all $A$, so if $epsilon > 0$ then this choice of $y$ works
to show that $y in A$ and $d(y,x) < epsilon$. Consequently, $x$ is a limit point of each $A$.

However, the union is not done correctly, and here's why : fix $epsilon_1 > 0$. What we know, is that $x$ is a limit point of $cup_A in F A$, so there is some $y in A_1 (in F)$ such that $d(y,x) < epsilon$. Note that for some other $epsilon_2$, there may be some other $A_2$, and for some other $epsilon_3$ some other $A_3$ : in short, it is possible that for all $A in F$, all $epsilon$ below some point stop working out after some time. That is, it is possible that for all $A in F$ , there exists $epsilon_A > 0$ such that for all $y in A$, we have $d(y,x) > epsilon_A$. So $x$ would not be a limit point of any of the $A$ but be a limit point of their union.

The best example of this, is a convergent sequence broken into parts : for example, let $A_n = frac 1n$ be singleton sets with the element $frac 1n$. Then, if you take the union of $A_n$, you get the set $1,frac 12,frac 13,...$ which has a limit point $0$. But $0$ is not a limit point of any $A_n$.

However, you can check that if $F$ is a finite set of sets, then actually this fact is true. See if you can figure out why.