Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2009 / 4

For closed sets in $\mathbb{R}$, the answers are $X_1$: no; $X_2$: yes; $X_3$: yes. The first interval omits its limit point zero. The second contains its finite limit points, and the empty set is closed in every topology.

For open sets in $\mathbb{R}$, the answers are $X_1$: yes; $X_2$: no; $X_3$: yes. Every point of the positive ray has a small interval contained in it, while zero is not an interior point of the nonnegative ray. The empty set is open by the topology axioms.

For open sets in $\mathbb{C}$, the answers are $X_1$: no; $X_2$: no; $X_3$: yes. Every complex disk centered at a real point contains nonreal points, so a nonempty subset of the real axis cannot be open in the complex plane. The empty set remains open.

The modulus map is continuous, the denominator $1+|z|$ is positive everywhere, and sums, products and quotients with nonzero denominator preserve continuity. Hence $f$ is continuous throughout $\mathbb{C}$.

For $g$, $|g(z)|=|z|$, so $g(z)\to 0=g(0)$ as $z\to 0$. The chosen argument is continuous locally away from the negative real axis, making $g$ continuous there as well. At a point $-r$, $r>0$, approach along $z=r{e}^{i\theta }$. From $\theta \uparrow \pi$, $g(z)\to ir$; from $\theta \downarrow -\pi$, $g(z)\to -ir$. The value assigned on the cut is $ir$. Thus $g$ is discontinuous at every strictly negative real point, and therefore is not a continuous map on the entire complex plane. Its radial factor is $r$, as printed, rather than the $\sqrt{r}$ of a square root.