I managed to prove that:

$$\displaystyle\bigcup_{i=1}^n A_i=A_1\cup(A_1^c\cap A_2)\cup(A_1^c\cap A_2^c\cap A_3)\cup\dots\cup(A_1^c\cap\dots\cap A_{n-1}^c\cap A_n)$$

for $\forall n \in\mathbb{N}$. Does this automatically proves that:

$$\displaystyle \bigcup_{n=1}^\infty A_n= \bigcup_{n=1}^\infty (A_1^c \cap\dots\cap A_{n-1}^c\cap A_n) $$

If not, what am I missing? Thanks for helping!


The second statement is true, and you can certainly use the first statement to prove it - but it's wise to be careful when you extend anything to infinity. You could notice that, as you have proved, for finite $k$ it holds that $$\bigcup_{n=1}^kA_n=\bigcup_{n=1}^k(A_1^c\cap \cdots\cap A_{n-1}^c\cap A_n)$$ then, since, when $k$ is infinite, we are just taking the union of all the above sets (that is the sets $\bigcup_{n=1}^kA_n$, not directly the $A_n$ this time), clearly the infinite unions are equal, since every member thereof is.

However, it'd probably just be easier to prove that, if $x$ is an element of $\bigcup_{n=1}^{\infty}A_n$, then there is some least $i$ such that $x\in A_i$ and then it follows that $x\in (A_1^c\cap \cdots \cap A_{i-1}^c\cap A_i)$ - which proves the theorem directly (for any sequence of sets $A_i$ with any well-ordered set of indices), and I imagine is similar to how you proved the finite case - so that might be a more elegant way to do it.

  • $\begingroup$ Thanks for your answer! I managed to prove the infinite case by using your 2nd suggestion (I couldn't understand the final bits of the 1st paragraph though). It was easier than expected to prove that each set is contained in the other one. @Meelo $\endgroup$ – Guilherme Salomé Nov 8 '14 at 4:05

$$\cup_{n=1}^{\infty}A_n=A_1 \cup (A_2 \setminus A_1) \cup (A_3 \setminus (A_1 \cup A_2) ) \cup \cdots \cup (A_n \setminus (A_1 \cup \cdots A_{n-1}) \cup \cdots= A_1 \cup (A_2 \cap A^c_1) \cup (A_3 \cap (A^c_1 \cap A^c_2) ) \cup \cdots \cup (A_n \cap (A^c_1 \cap \cdots \cap A^c_{n-1})) \cup \cdots$$


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