Question

\(A \subset X\) and consider \( \rm T=2^A \cup \{X\} \)

Show that union of an infinite indexed family in T is also in T.

Answer 1

@ChristopherToni

Answer 2

I assume we're talking about a "countable" infinite union?

Answer 3

I think so... it is given \(X\ne \varnothing \)

Answer 4

\(2^A\) is power set of A

Answer 5

I think the crux of the argument is going to rely on the fact that \(2^A= \mathcal{P}(A)=\{\text{all subsets of }A\}\); give me a few to see if I can set something up. XD

Answer 6

thank you :) so far i have below :
1) \(\varnothing \in T \)
2) \((P, Q) \in T \implies P \cup Q \in T\)

Answer 7

i have already proven above two and got stuck on main question because i am not so sure what they mean by "infinite indexed family in T"

Answer 8

Usually, they mean a countable indexed family; say for example \(\{U_i\}\) is a countable collection of sets in \(T\). Then you would want to show that \(\displaystyle \bigcup\limits_{i=1}^{\infty}U_i \subset T\). Another way to say it then would be to let \(N\) be a countable index set and let \(\alpha \in N\). Then define the collection of sets by \(\{U_{\alpha}\}\) and show that \(\displaystyle \bigcup_{\alpha \in N}U_{\alpha}\subset T\).

Answer 9

I don't see why you would assume its countable.

Answer 10

But any infinite set, can be partitioned into a disjoint union of countable sets.

Answer 11

\(\displaystyle \bigcup\limits_{i=1}^{\infty}U_i \subset T \)

Is \(U_i\) an element of \(T\) ? If so, would the upper index stop at the size of \(T\) ?

Answer 12

X = {1, 2, 3, 4, 5, 6}
A = {2}

T = {phi, {2}, X}

Answer 13

Hm. I suppose it would be a finite union in the event that \(X\) is finite. However, if \(X\) isn't finite, then \(T\) isn't finite either.

Answer 14

And then it would probably make more sense to talk about infinite (countable) unions.

Answer 15

Okay thatmakes sense thanks !

I am starting like below :
\(P \in \displaystyle \bigcup\limits_{i=1}^{\infty}U_i \implies P\in \text{some } U_j \in T~ \text{ for some j} \)

\(Q \in \displaystyle \bigcup\limits_{i=1}^{\infty}U_i \implies Q\in \text{some } U_k \in ~T\text{ for some k} \)

Answer 16

not getting ideas on how to proceed further

Answer 17

It is pretty easy. Consider two cases:
Case 1; All elements involved in the union are distinct from X. Hence they all all subsets of A. Hence their union is a subset of A. Hence belong to T.
Case 2. At least one of the elements involved in the union is X. In this case, the union is
X. Hence belongs to T

Answer 18

Wow! that looks neat thanks a lot :D

Answer 19

\(P \in \displaystyle \bigcup\limits_{i=1}^{\infty}U_i \implies P\in \text{some } U_j \in T~ \text{ for some j} \)

\(Q \in \displaystyle \bigcup\limits_{i=1}^{\infty}U_i \implies Q\in \text{some } U_k \in ~T\text{ for some k} \)

\(P \ne X ~ \land ~ Q\ne X \implies P \cup Q \subset A \in T\)
\(P = X ~ \lor ~ Q= X \implies P \cup Q = X \in T ~~\square \)

thanks @eliassaab

Answer 20

that doesnt sound like infinite indexed set

Answer 21

\( \left \{ V_\alpha :\forall \alpha \in \Delta \right \} \in T\\
V_\alpha - A_=\varnothing ,\forall \alpha \in \Delta \\
V_\alpha \cap A^c=\varnothing ,\forall \alpha \in \Delta \\
\\

\begin{matrix}
V_1 \cap A^c=\varnothing \\
V_2 \cap A^c=\varnothing \\
....
\\
\bigcup_{\alpha \in \Delta} (V_\alpha \cap A^c)= \varnothing\\

(\bigcup_{\alpha \in \Delta} V_\alpha) \cap A^c= \varnothing\\
(\bigcup_{\alpha \in \Delta} V_\alpha) - A= \varnothing\\ \text{thus } \bigcup_{\alpha \in \Delta} V_\alpha \in T
\end{matrix} \)