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Suppose $\mathscr{A}=\{A_i:i\in\mathbb{N}\}$ is a family of sets such that for all $i,j\in\mathbb{N}$, if $i\le j$ then $A_j\subseteq A_i$. (Such a family is called a nested family of sets).

a) Prove for every $k\in\mathbb{N}$, $\bigcap_{i=1}^{k}A_i=A_k$

b) Prove that $\bigcup_{i=1}^{\infty}A_i=A_1$

Proof of a):

Suppose $x\in\bigcap_{i=1}^{k}A_i$

iff $x\in A_i$ for all $i\in\mathbb{N}$.

iff $x\in A_j$ since $A_i\cap A_j= A_j$, $A_j \subseteq A_i$ and $A_i=A_j$ since $j\in\mathbb{N}$

iff $x\in A_k$ for all $k\in\mathbb{N}$, since $A_k=A_j\cap A_k$, $A_k\subseteq A_j$ and $k\in\mathbb{N}$.

Proof of b)

Suppose $x\in\bigcup_{i=1}^{\infty}A_i$. Then $x\in A_1$ since it is contained in $\bigcup_{i=1}^{\infty}A_i$. Hence $\bigcup_{i=1}^{\infty}A_i\subseteq A_1$.

Now suppose $x\in A_1$ since $1\le i \le j$ for all $i,j\in\mathbb{N}$, it follows $1\le i \le j$ for some $i,j\in\mathbb{N}$. Since $A_j\subseteq A_i\subseteq A_1$, it follows $x\in A_i$ for some $i\in\mathbb{N}$. Hence $x\in > \bigcup\limits_{i\in\mathbb{N}}A_i$. Hence $x\in\bigcup_{i=1}^{\infty}A_i$.

Hence $\bigcup_{i=1}^{\infty}A_i=A_1$

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3 Answers 3

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I’m afraid that neither argument makes sense. In the first argument it is not true that $x\in\bigcap_{i=1}^kA_i$ iff $x\in A_i$ for all $i\in\Bbb N$. For example, let $A_1=\{0\}$, and let $A_i=\varnothing$ for all $i\ge 2$; then $\{A_i:i\in\Bbb N\}$ is a nest. Now take $k=1$: then

$$\bigcap_{i=1}^kA_i=\bigcap_{i=1}^1A_i=A_1=\{0\}\,,$$

so $0\in\bigcap_{i=1}^kA_i$, but clearly $0\notin A_2$ (and in fact $0\notin A_i$ whenever $i\ge 2$).

In the next line you talk about some $A_j$, but you never define it, so nothing that you say about it is meaningful. At the end you conclude that $x\in A_k$ for all $k\in\Bbb N$; this is just a repetition of what you said in the second line, but with a different name for the subscript, and it is every bit as false in general.

In the argument for (b) you start with an $x\in\bigcup_{i=1}^\infty A_i$ and conclude that it must be in $A_1$, since $A_1\subseteq\bigcup_{i=1}^\infty A_i$. Your reasoning here is exactly backwards: the fact that $A_1\subseteq\bigcup_{i=1}^\infty A_I$ tells you that if $x\in A_1$, then $x\in\bigcup_{i=1}^\infty A_i$, not the reverse.

Then when you try to prove the trivial implication that if $x\in A_i$, then $x\in\bigcup_{i=1}^\infty A_i$, you pick some unspecified positive integers $i$ and $j$ with $i\le j$ and claim that the fact that $A_j\subseteq A_i\subseteq A_1$ somehow proves that $x\in A_i$ for some $i\in\Bbb N$. This is a complete non sequitur, and in any case we already know that $x\in A_i$ for some $i\in\Bbb N$, since we assumed at the beginning that $x\in A_1$, and certainly $1\in\Bbb N$.

To prove (a), first observe that if $x\in\bigcap_{i=1}^kA_i$, then $x\in A_k$, so $\bigcap_{i=1}^kA_i\subseteq A_k$. Now suppose that $x\in A_k$, and let $i\in\{1,\ldots,k\}$; then $i\le k$, so $A_k\subseteq A_i$, and therefore $x\in A_i$. Thus, $x\in A_i$ for all $i\in\{1,\ldots,k\}$, and hence $x\in\bigcap_{i=1}^kA_i$. In other words, $A_k\subseteq\bigcap_{i=1}^kA_i$, and it follows that $\bigcap_{i=1}^kA_i=A_k$.

For (b) it’s immediate that if $x\in A_1$, then $x\in\bigcup_{i=1}^\infty A_i$, so $A_1\subseteq\bigcup_{i=1}^\infty A_i$. Now suppose that $x\in\bigcup_{i=1}^\infty A_i$; then there is some $i\in\Bbb N$ such that $x\in A_i$. And $i\ge 1$, so $A_i\subseteq A_1$, so $x\in A_1$. Thus, $\bigcup_{i=1}^\infty A_i\subseteq A_1$, and we conclude that $\bigcup_{i=1}^\infty A_i=A_1$.

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None of the proofs is correct.

a) You mention a $j$ without saying what $j$ is.

b) You claim that every element of $\bigcup_{i\in\Bbb N}A_i$ belongs to $A_1$. That's not obvious.

These statements can be proved as follows:

a) Clearly, $\bigcap_{i=1}^kA_i\subset A_k$. On the other hand, if $x\in A_k$ and if $i\in\{1,2,\ldots,k\}$, then, since $i\leqslant k$, $x\in A_i$. Since this occurs for each $i\in\{1,2,\ldots,k\}$, $x\in\bigcap_{i=1}^kA_i$.

b) Clearly, $A_1\subset\bigcup_{i\in\Bbb N}A_i$. On the other hand, if $x\in\bigcup_{i\in\Bbb N}A_i$, then $x\in A_i$, for some $i\in\Bbb N$. Since $i\geqslant1$, $x\in A_1$.

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I think it makes more sense to call it a 'decreasing family of sets' than a 'nested family of sets' because you can consider all those sets as subsets of $A = \bigcup_{i \in \mathbb N} A_i$, and the power set $\mathcal P(A)$ can be equipped with a partial order where $B \le C$ if and only if $B \subseteq C$. This makes the sequence $\mathscr A$ decreasing in that sense, since $i \le j$ implies $A_i \ge A_j$.

Your proof of (a) shows that you are confused about what the notation actually means. You are right that since $A_i \supseteq A_k$, $x \in \bigcap_{i=1}^k A_i$ if and only if $x \in A_k$, but the formulation is not quite enlightening. Usually you use the axiom of extensionality and try to prove that both sets admit the same elements. Something like this.

($\supseteq$) Since $1 \le i \le k$, we have $A_i \supseteq A_k$ for $i \in \mathbb N$ satisfying $1 \le i \le k$, and therefore $\bigcap_{i=1}^k A_i \supseteq A_k$.

($\subseteq$) We have $\bigcap_{i=1}^k A_i \subseteq A_k$ because $A_k$ is a member of that intersection.

Since both sets contain the same elements, they are equal. Usually the above two sentences suffice, i.e. proving that ($\subseteq$) and ($\supseteq$) hold.

You can do the proof for (b) in the same way. Again, not quite sure what you were trying to say, you seem confused there as well.

($\subseteq$) We have $A_1 \supseteq A_i$ for $i \in \mathbb N$ since $i \ge 1$, and therefore $\bigcup_{i=1}^{\infty} A_i \subseteq A_1$.

($\supseteq$) We have $\bigcup_{i=1}^{\infty} A_i \supseteq A_1$ because $A_1$ is a member of that union.

Hope that helps,

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