Topology Exercise - Solution Verification

Question

Hi everyone I've got this long topology exercise where I'm not completely sure about each one of my answers.

Let $\mathbb{D}:= \{ (x,y) \in \mathbb{R}^2: x^2+y^2\leq 1 \}$ and $\tau$ be the family of open sets thus defined $$A\in \tau \iff \rho e^{i\theta} \in A \implies \rho e^{i\theta +i\pi/2} \in A$$

1. Prove $\tau$ is a topology on $\mathbb{D}$ and compare it with Euclidean topology

I verified that it is indeed a topology. Regarding the comparison with the Euclidean topology, I think that this two topologies aren't comparable. In $\tau$ there are open sets like $A= \big\{ \frac{1}{2}e^{i\pi/4}, \frac{1}{2}e^{i\pi5/4}\big\}$, which is not open in Euclidean topology. Vice versa I can find a small open ball in Euclidean topology that contains $\rho e^{i\theta}$ but not contains $\rho e^{i\theta +i\pi/2}$, so they aren't comparable.

2. Closure and interior of $[-1,1]\times\{0\}$

As I argued in the previous point $\tau$ has open sets containing only two points. So the interior is reduced to the point $O=(0,0)$ because every other point on the segment from $-1$ to $1$ on the x-axes has a "counterpart" on the y-axes. For the same reason I think the closure is the cross $([-1,1]\times\{0\}) \cup( \{0\}\times[-1,1])$

3. Let $Y := [-1,1]\times\{0\} $ with induced subspace topology, determine if $p:X\to Y , p(\rho e^{i\theta})=(\rho\cos(\theta),0))$ is a continuous function and/or open

The first thing I thought of was that on $Y$ we got the discrete topology and after calculating the preimages of a generic point of $Y$ I think that this function is neither continuous neither open. Thank you very much for your time and help!

Answer 1

I freely mix $\Bbb R^2$-notation and $\Bbb C$-notation for convenience.

Your choice of $A$ isn't open in $\tau$. However, the very similar $A:=\{1,i,-i,-1\}$ is open in $\tau$ without being Euclidean-open. So, you're right the topologies are incomparable.

$\tau$ does not have open sets containing only two points. Elements of $\tau$ are either empty, are $\{0\}$, have a finite multiple of four points, or are infinite.

You're correct, for the correct reasons, that $\{0\}$ is the interior of $[-1,1]\times\{0\}$ in $\tau$. "For the same reason" you think the closure is the cross? I don't understand that choice. The cross is a priori the smallest open set that contains $[-1,1]\times\{0\}$, and a priori has nothing to do with its closure. However, you are still right that this set is the closure since a limit point $z$ of $[-1,1]\times\{0\}$ must have $i^kz\in[-1,1]\times\{0\}$ for some integer $k$. Indeed, $X$ is a space where every open set is also closed.

$Y$ does not inherit the discrete topology. $\{0\}$ is open in $Y$ but that's the only isolated point; given $x\in Y$ there is a smallest neighbourhood of $x$ in $Y$, namely $\{x,-x\}$.

It's asked if the projection $X\to Y$ is continuous and or open. You haven't shown us your thoughts too clearly on this one, but I invite you to reconsider. It is not continuous but it is open.