For the quotient map $q:X\to X/\sim$, the quotient topology is defined byTake the Hausdorff space $X=\mathbb{R}$ and identify two numbers precisely when they are both rational or both irrational. The quotient has two points. The inverse images of its singleton subsets are $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$, neither of which is open in $\mathbb{R}$, since each is dense and has dense complement. Hence the quotient has the indiscrete topology and is not Hausdorff.

The product topology on $X\times Y$ has as a basis of a topology the rectangles $U\times V$ with $U$ open in $X$ and $V$ open in $Y$; arbitrary unions of such rectangles are its open sets.

For distinct points $(x_1,y_1),(x_2,y_2)$, at least one coordinate differs. If $x_1\ne x_2$, disjoint open neighbourhoods $U_1,U_2$ in the Hausdorff space $X$ give disjoint open neighbourhoods $U_1\times Y,U_2\times Y$ of the two points. If the second coordinates differ instead, use $X\times V_1,X\times V_2$. Thus a product of two Hausdorff spaces is Hausdorff.

A sequence $x_n$ converges to $x$ if every open neighbourhood of $x$ contains all sufficiently late terms. A Hausdorff space separates any two distinct points by disjoint open neighbourhoods.

If a sequence had distinct limits $x,y$ in a Hausdorff space, choose disjoint open neighbourhoods $U$ of $x$ and $V$ of $y$. Convergence to each limit implies that all sufficiently late terms lie in both $U$ and $V$, contradicting $U\cap V=\varnothing$. Hence a convergent sequence in a Hausdorff space has a unique limit.

Let $d=h-g$. It vanishes outside the null set $E$, so part (e) proves $d$ is a measurable function, regardless of its values on $E$. Since the sum of two measurable functions is measurable, $h=g+d$ is measurable. This is the completeness of Lebesgue measure under changes on null sets.

If $a\ge 0$, the set $\{f>a\}$ is contained in $E$, since $f=0$ outside $E$. It is therefore a null set and is Lebesgue measurable. If $a<0$, its complement $\{f\le a\}$ is contained in $E$ and is measurable for the same reason. Complements of measurable sets are measurable. Thus every upper level set is measurable, proving $f$ is a measurable function.

A real measurable function is one whose inverse image of every open subset of $\mathbb{R}$ is Lebesgue measurable. Equivalently,These upper rays generate the Borel sigma-algebra, making this equivalent to requiring measurable inverse images of all Borel sets.

If $A\subseteq E$ and $E$ is a null set, every interval cover of $E$ also covers $A$. Thus monotonicity of Lebesgue outer measure gives $0\le m^{\ast }(A)\le m^{\ast }(E)=0$. Hence every subset of a null set is null, and in particular is Lebesgue measurable.

A null set has Lebesgue outer measure zero. Equivalently, for every $\epsilon >0$ it can be covered by countably many open intervals whose total length is less than $\epsilon$. Such a set is automatically Lebesgue measurable and has Lebesgue measure zero.

The Lebesgue outer measure isThe Lebesgue inner measure is $m_{\ast }(E)=\sup \{m(K):K\subseteq E,K\text{ compact}\}$, with $m_{\ast }(\varnothing )=0$. For a bounded set inside a finite interval $I$, this is equivalently $|I|-m^{\ast }(I\setminus E)$.

A set is Lebesgue measurable if it satisfies Carathéodory's criterion:For finite outer measure this is equivalent to equality of inner and outer measure. For a general unbounded set, an equivalent local criterion is $m_{\ast }(E\cap [-n,n])=m^{\ast }(E\cap [-n,n])$ for every positive integer $n$. Merely writing $m_{\ast }(E)=m^{\ast }(E)=\infty$ would not establish measurability.

Realize the cube's vertices as $(\pm 1,\pm 1,\pm 1)$ and take the four with coordinate product $+1$. They form a regular tetrahedron, and their negatives form the other one. Every proper rotation of the first tetrahedron also permutes the negatives, hence is a cube rotation. This embeds $G_1$ as a subgroup of $G_2$.

The embedded group preserves each of these two four-vertex sets and acts transitively on each. Thus the resulting action on all eight cube vertices has two orbits, each of size four.

For a regular tetrahedron, rotations fixing a vertex rotate about the line from that vertex through the opposite face centre. Their angles are $0,2\pi /3,4\pi /3$, so the stabilizer is a cyclic group of order three. Rotations act transitively on the four vertices, giving $|G_1|=4\cdot 3=12$ by the orbit-stabilizer theorem.

For a cube, the corresponding axis is the body diagonal through the fixed vertex and its opposite. The three incident edges can be cyclically permuted by the same three angles, so again the stabilizer is cyclic of order three. The orbit has eight vertices. Hence

For a group action, the orbit and stabilizer areThe orbit-stabilizer theorem for a finite group states

Suppose a rational function $P(t)/Q(t)\in {\mathbb{F}}_p(t)$, with nonzero polynomials $P,Q$, had $p$th power $t$. Then $P(t{)}^p=tQ(t{)}^p$, so comparison of degrees would give $p\mathrm{deg}P=1+p\mathrm{deg}Q$, impossible modulo $p$. Thus $X^p-t$ has no root in $K$, and part (b) proves it is irreducible.

In a splitting field containing $\alpha$ with ${\alpha }^p=t$,It has one distinct root with multiplicity $p$, so it is inseparable. This is the purely inseparable polynomial over a rational function field.