For Problem Q-2:

The question is whether each example describes a partition. Reasons weren't required for positive answers, but I'll include them.

(a) No, because the problem says any two blocks are to have a plant in common rather than to be pairwise disjoint. The union of the blocks is the whole set, though.

(b) Yes, because the planes together fill up $\mathbb{R}^3$ and yet any two are disjoint (since they're parallel).

(c) Yes, like (b).

(d) No. The union of the 1-dimensional subspaces is the whole space, but two 1-dimensional subspaces intersect in $\{$0$\}$ rather than the empty set.

(e) Yes; the sets of even integers and odd integers are disjoint and their union is all integers.

(f) Yes; any two blocks are disjoint and any integer is in one of the blocks.

(g) Yes; the union of the singleton subsets is the whole space and any two distinct singleton subsets have empty intersection.

(h) Yes. Recall that $f^{-1}(t)$ really means $f^{-1}(\{t\})$. The union of these inverse images is all of $S$ since $x \in f^{-1}(y)$ for $y = f(x)$. The inverse images are disjoint since each $x \in S$ goes to only one element of $T$ by the definition of a function.

Notice that we could be fancier and say that we are first partitioning the image of $f$ into singletons as in (g) and then using rules (iii) and (iv) of Problem O-4.

Also notice that (b) and (c) can be viewed as an application of (h) in any particular example if we can find a linear transformation on $S$ whose null space is the 1- or 2-dimensional subspace.

(i) No; blocks $A _ r$ and $A _ s$ (with $r \neq s$) are not disjoint since the constant function 0 is in both. There are other functionsn too, such as $(x-r)(x-s)$.

(j) Yes, this is a ``tiling of the plane''. Notice that it's important to use ``half-open intervals'' so the tiles fit together with no overlap and no holes between.

Note: An example not included that could have been is

(k) There is just one block, consisting of all of $S$. This does count as a partition.

For Problem Q-3:

(a) The union of the nonzero parts of 1-dimensional subspaces is the nonzero part of the vector space, since every nonzero vector is in the nonzero part of the 1-dimensional subspace it generates.

Any two 1-dimensional subspaces are disjoint except for 0, since if a nonzero vector $v$ is in both, then both subspaces consist of the nonzero scalar multiples of $v$ and so are the same.

(b) Let $F =$   GF$(q)$. Then $V \cong F^3$ and so $V$ has $q ^3$ elements, so the nonzero part of $V$ has $q^3 -1$ elements. A 1-dimensional subspace is isomorphic to $F^1 = F$ and so has $q$ elements; its nonzero part has $q-1$ elements. Since $V \setminus \{$0$\}$ is partitioned as in (a) and the blocks have equal size, the number of blocks is $(q^3 -1)/(q-1) = q^2 + q + 1$. (If you don't know how to factor $q^3 -1$ as a polynomial in $q$, use long division of polynomials.)

For the case $q = 2$ this gives 7, which checks with the answer to Problem G-4.