# Collections of Sets, Problem 2

### Solution part (a)

f(1) = {b,c,e,f}

f(3) = {d}

### Self-check

Do you have the same number of layers of curly brackets in your value for
f(3)?

### Solution part (b)

Notice that
P = {f(1), f(2), f(3)} = {{b,c,e,f}, {a,g}, {d}}.

P obviously doesn't contain the empty set, so that partition property holds.

The elements of P don't overlap, so that partition property is also ok.

However, none of the elements of P contains the node X. So P doesn't cover
all the elements of V. So that partition property isn't satisfied.

So P isn't a partition of V.