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I'm trying to prove the following:

" Suppose that one has proven the proposition that if $A \subseteq B$ and $C \subseteq D$, then $A \cup C \subseteq B \subseteq D$. Prove that for any integer $n \geq 2$ that if sets $A_1, A_2,...,A_n$ and $B_1, B_2,...B_n$ are sets that satisfy $A_j \subseteq B_j$ for $j = 1, 2, ..., n$ then $$\bigcup_{j=1}^n A_j\subseteq \bigcup_{j=1}^n B_j."$$

I'm not sure if I what I came up with makes sense logically, and would appreciate some feedback.

Proof:

Define P(n): $\bigcup_{j=1}^n A_j\subseteq \bigcup_{j=1}^n B_j$.

Base case $(n=2)$:

$\bigcup_{j=1}^2 A_j \subseteq \bigcup_{j=1}^2 B_j = A_1 \bigcup A_2 \subseteq B_1 \bigcup B_2$. So $P(2)$ holds.

Inductive step: Assume $P(k)$ holds for some $k \geq 2$. So $$\bigcup_{j=1}^k A_j \subseteq \bigcup_{j=1}^k B_j = A_1 \bigcup A_2 \bigcup ... \bigcup A_k \subseteq B_1 \bigcup B_2 \bigcup ...\bigcup B_k.$$ Notice, $$\bigcup_{j=1}^{k+1} A_j \subseteq \bigcup_{j=1}^{k+1} B_j = A_1 \bigcup A_2 \bigcup ... \bigcup A_k \bigcup A_{k+1} \subseteq B_1 \bigcup B_2 \bigcup ... \bigcup B_k \bigcup B_{k+1}.$$ So $P(k+1)$ holds and by induction $P(n)$ holds for all $n \geq 2$.

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  • $\begingroup$ Please do not rely on images to convey key information about your question. See here for an explanation of why it is frowned upon. $\endgroup$ Commented Oct 18, 2021 at 1:39
  • $\begingroup$ @ArturoMagidin Oh, I didn't realize this was an issue! Should I edit it, then? $\endgroup$ Commented Oct 18, 2021 at 1:40
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    $\begingroup$ Yes. Please replace the image with text; you can indent it (using >) to indicate you are quoting. $\endgroup$ Commented Oct 18, 2021 at 1:41
  • $\begingroup$ You have written out what the claims involve, but you haven’t really shown them to be true. For example, for the base case, you need to show that if $A_1 \subseteq B_1$, and $A_2 \subseteq B_2$, then $A_1\cup A_2 \subseteq B_1\cup B_2$ $\endgroup$
    – Bram28
    Commented Oct 18, 2021 at 1:41
  • $\begingroup$ @ArturoMagidin Ok, thank you. $\endgroup$ Commented Oct 18, 2021 at 1:41

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Thanks for your question.

I will continue from inductive step.

Inductive step: Assume $P(k)$, then we want to show it holds for the inductive step $P(k+1)$:

$$\bigcup_{j=1}^{k+1} A_j \subseteq \bigcup_{j=1}^{k+1} B_j = \left(A_1 \bigcup A_2 \bigcup ... \bigcup A_k\right) \bigcup A_{k+1} \subseteq \left( B_1 \bigcup B_2 \bigcup ... \bigcup B_k\right) \bigcup B_{k+1}.$$

You can then consider the two paraenthsized groups as one group and thus you can consider them as 2 elements similar to how you did with base case, which will give you final result.

Please let me know if anything is not clear.

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