# Analysis help->Suppose that g1 : A1 → A2 and g2 : A2 → A3. If B ⊂ A3, show that:

• cooljosh2k2
In summary: B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)k (B)))), which is the statement for n=k+1. Therefore, by induction, the statement holds for all n≥2.
cooljosh2k2
Analysis help-->Suppose that g1 : A1 → A2 and g2 : A2 → A3. If B ⊂ A3, show that:

## Homework Statement

Suppose that g1 : A1 → A2 and g2 : A2 → A3. If B ⊂ A3, show that
a) (g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (B)).

b) Now suppose that n ≥ 2 and gi : Ai → Ai+1 for i = 1, 2, . . . n. If B ⊂ An+1 show that
(gn ◦ gn−1 ◦ . . . ◦ g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)n (B))))

2. The attempt at a solution

a) first, g^(-1)2(B) = {x$$\in$$A2 : g2(x)$$\in$$B}

--> g^(-1)(g^(-1)2(B)) = g^(-1)1({x $$\in$$ A2 : g2(x) $$\in$$ B})

--> g^(-1)1({x $$\in$$ A2 : g2(x) $$\in$$ B}) = {y $$\in$$ A1 : g1(y) $$\in$$ {x $$\in$$ A2 : g2(x) $$\in$$ B}}

-->g^(-1)1(g^(-1)2(B)) = {y $$\in$$ A1 : (g2 ◦ g1)^(-1)(y) $$\in$$ B}

Which by definition = ((g2 ◦ g1)^(−1)(B)

b) I am not sure how to do part B, can anybody help me. Would i need to use induction? I apologize if this all looks confusing, i tried to make it as clear as possible.

Thanks

Homework Equations

N/A

The Attempt at a Solution

b) Yes, you will need to use induction. The key idea is to show that for n=2, the statement is true, and then assume that it holds for n=k and prove that it holds for n=k+1. This will show that the statement holds for all n≥2.

For n=2, we have the statement (g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (B)), which we have already shown to be true in part a.

Now, assume that the statement holds for n=k, i.e. (gk ◦ gk−1 ◦ . . . ◦ g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)k (B)))). Now, we want to prove that it holds for n=k+1. Let's call gk+1=g and Ak+1=A. Then we have gk+1 : Ak → Ak+1. Now, we can rewrite the statement as (g ◦ gk ◦ gk−1 ◦ . . . ◦ g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)k (B)))). Using the associative property of function composition, we can rewrite this as (gk+1 ◦ gk ◦ gk−1 ◦ . . . ◦ g2 ◦ g1)^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)k (B)))). Now, using the induction hypothesis, we can rewrite this as (gk+1 ◦ (gk ◦ gk−1 ◦ . . . ◦ g2 ◦ g1))^(−1)(B) = g^(−1)1 (g^(−1)2 (. . . (g^(−1)k (B)))). But this is just (gk+1 ◦ (gk ◦ gk−1 ◦ . . . ◦

## What is the definition of g1 and g2?

g1 and g2 are functions that map elements from one set to another. In this case, g1 maps elements from set A1 to set A2 and g2 maps elements from set A2 to set A3.

## What does B ⊂ A3 mean?

B ⊂ A3 means that set B is a subset of set A3. This means that all elements in set B are also present in set A3, but set A3 may contain additional elements that are not in set B.

## How do I show that B ⊂ A3?

To show that B ⊂ A3, we must prove that all elements in set B are also present in set A3. This can be done by taking an arbitrary element from set B and showing that it is also present in set A3.

## How does g1 and g2 relate to B ⊂ A3?

Since g1 maps elements from set A1 to set A2 and g2 maps elements from set A2 to set A3, we can use these functions to show that elements in set B are also present in set A3. This is because g1 and g2 can be used to map elements from set A1 to set A3, which includes all elements in set B.

## Can I use g1 and g2 to prove B ⊂ A3 for any set B and A3?

No, g1 and g2 can only be used to prove B ⊂ A3 if A1, A2, and A3 are specific sets. If the sets are different, then g1 and g2 may have different mappings and cannot be used to prove B ⊂ A3.

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