Welcome to our community

Be a part of something great, join today!

intersection and union

dwsmith

Well-known member
Feb 1, 2012
1,673
Let $F$ be a collection of sets in $\mathbb{R}^n$, and let $S = \bigcup\limits_{A\in F}A$ and $T = \bigcap\limits_{A\in F}A$.
For each of the following statements, either give a proof of exhibit a counterexample.


If $x$ is an accumulation point of $T$, then $x$ is an accumulation point of each set $A$ in $F$.
False. $T$ could be the set $\{x\}$ and $x$ may only be a point in some of $A_n$. Correct?

If $x$ is an accumulation point of $S$, then $x$ is an accumulation point of at least one set $A$ in $F$.
Not sure but wouldn't $x$ have to be an accumulation point of all $A$?
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Let $F$ be a collection of sets in $\mathbb{R}^n$, and let $S = \bigcup\limits_{A\in F}A$ and $T = \bigcap\limits_{A\in F}A$.
For each of the following statements, either give a proof of exhibit a counterexample.

If $x$ is an accumulation point of $T$, then $x$ is an accumulation point of each set $A$ in $F$.
False. $T$ could be the set $\{x\}$ and $x$ may only be a point in some of $A_n$. Correct?

If $x$ is an accumulation point of $S$, then $x$ is an accumulation point of at least one set $A$ in $F$.
Not sure but wouldn' $x$ have to be an accumulation point of all $A$?
In $\mathbb{R}^n$ no finite set can have an accumulation point.
In the first part $T\subseteq A$ for all $A\in F$. So if $T$ has an accumulation point then each $A$ must be infinite.

If $O$ an open set and $x\in O$ then $\exists t\in O\cap (T\setminus \{x\})$.
Is that also true for each $A~?$