# Spooky geometry V

The complement of an algebraic set in $A^n(\Bbb{C})$ is path-connected.

# Contractible manifolds

Classify the closed contractible manifolds (Recall that a manifold is closed if it is compact and its boundary is empty.)

# Connect the dots

Construct a countable, connected topological space with at least two points, satisfying the highest separability axiom you can.

Notice that if two points can be separated by a continuous function, then a connected space with more than one point is uncountable, so your space cannot be $T_{3 \frac 12}$ (or higher).

# No swapping

There is no continuous function $f:\Bbb{R}\to\Bbb{R}$ such that $f(\Bbb{Q})\subseteq \Bbb{R}\setminus\Bbb{Q}$ and $f(\Bbb{R}\setminus\Bbb{Q}) \subseteq \Bbb{Q}$.

# Fixed points of the ball

Brouwer’s theorem shows that any map of the unit ball to itself has a fixed point. Evidently the collection of fixed points is a closed subset of the ball. What closed subsets are realized as the fixed point set of a continuous map of the ball into itself?

# Complements of closed sets

Construct  two homeomorphic closed subsets $A,B$ of some Euclidean space such that their complements are not homeomorphic, but show that if $A,B$ are closed homeomorphic subsets of $\Bbb R^{n+m}$ such that $A$ lies in $\Bbb R^n\times 0$ and $B$ lies in $0\times\Bbb R^m$, their complements in $\Bbb R^{n+m}$ are homeomorphic.

# Venn diagrams

Venn diagrams are useful since they represent every possible intersection between the sets that one is studying. Suppose that we want to draw a Venn diagram for a single set. Then it is enough to draw it in zero dimensions (the zero-dimensional space is just the space with one point): the set is represented by the single point in the zero-dimensional space. If we want to draw a Venn diagram that represents two sets we notice that the zero-dimensional space won’t do, there is not enough space. But in the one-dimensional space (the real line) we can draw both sets $A$ and $B$ in such a way that we have distinct regions for $A \setminus B$, $B \setminus A$ and $A \cap B$. Once again if we want to draw one more set, one dimension won’t do. But we can draw the classical Venn diagram in two dimensions.

First exercise: Show that we cannot draw a Venn diagram in two dimension for four sets using only circles.

Second exercise: Show that this can be done if we use for example rectangles.

Third exercise: Can we draw a Venn diagram for four sets in two dimensions in such a way that the region that represents every intersection is connected? And simply connected?