Composition of Relations (functions)

Question

Working with functions, more specifically function composition. In my case, for starters,

We have an island with the places haven, dale, sun, and ness.

We have a smaller island with the places east, bent and big.

And a tiny island with the place smallest.

We have relations between different places represented like Relation ⊆ place × place

Place ={Little, east, bent, big, ness, haven, sun, dale}

The relation is represented like (this is how the places are connected/related by roads etc)

Relation ={(haven, dale),(haven, sun),(haven, ness),(dale, sun),(east, bent),(east, big)}

The islands are connected by ferry- represented like

Ferry ⊆ place × place

Specifically:

Ferry ={(ness, smallest),(smallest, haven),(haven, ness)}

Question: what is the result/way to solve

1. Relation ∘ Ferry

2. Ferry ∘ Relation

3. Ferry−1∘ Relation

I have read up on composition of relations online and in books, but I don't understand it and I am stuck on how to solve these tasks/what the correct answer is. If someone could show me the correct answer in a simple way that would be highly appreciated.

Answer 1

Guide.

For finding Relation$\circ$Ferry start looking for an ordered pair $(a,b)$ in Ferry such that for some $c$ we have $(b,c)$ in Relation. That gives three questions associated with the three pairs in Ferry:

• Do we have a pair $(\text{smallest},\dots)$ in Relation? No!

• Do we have a pair $(\text{haven},\dots)$ in Relation? Yes! We can substitute $\text{dale}$, $\text{sun}$ and $\text{ness}$ for $\dots$ and the conclusion is that $(\text{smallest},\text{dale})$, $(\text{smallest},\text{sun})$ and $(\text{smallest},\text{ness})$ are elements of Relation$\circ$Ferry.

• Do we have a pair $(\text{ness},\dots)$ in Relation? No!

Now we found: Relation$\circ$Ferry$=\{(\text{smallest},\text{dale}),(\text{smallest},\text{sun}),(\text{smallest},\text{ness})\}$

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edit:

For finding Ferry$\circ$Relation start looking for an ordered pair $(a,b)$ in Relation such that for some $c$ we have $(b,c)$ in Ferry. That gives six questions associated with the six pairs in Relation:

• Do we have a pair $(\text{dale},\dots)$ in Ferrry? No!
• Do we have a pair $(\text{sun},\dots)$ in Ferry? No!
• Do we have a pair $(\text{ness},\dots)$ in Ferry? Yes! We can substitute $\text{smallest}$ for $\dots$ and the conclusion is that $(\text{haven},\text{smallest})$ is an elements of Ferry$\circ$Relation.
• Do we have a pair $(\text{sun},\dots)$ in Ferry? No!
• Do we have a pair $(\text{bent},\dots)$ in Ferry? No!
• Do we have a pair $(\text{big},\dots)$ in Ferry? No!

Now we found: Ferry$\circ$Relation$=\{(\text{haven},\text{smallest})\}$

For finding Ferry$^{-1}\circ$Relation do exactly the same thing but now not for Ferry but for Ferry$^{-1}$ which is the relation $\{(\text{smallest},\text{ness}),(\text{haven},\text{smallest}),(\text{ness},\text{haven})\}$