1. Let T be the set {a, b, c}. For each of the following topologies on T, determine whether or not T is connected and whether or not T is Hausdorff:
- ∅, {a}, {a,b}, {a,b,c}
- ∅, {a}, {b,c}, {a,b,c}
- ∅, {a}, {a,b}, {a,c}, {c}, {a,b,c}

2. Determine whether or not the half-open interval [0, 1/2) is compact. If it is, give a proof. If not, give an example of an unbounded function [0, 1/2) → R

3. Prove that if T is a Hausdorff space and x₁...xₙ is a finite list of distinct points in T, then there are open sets U₁...Uₙ each containing one and only one of the points x₁...xₙ.

4. Verify that the open sets on the real line with a double point, given in the example 1* do actually form a topology.

5. Prove that the stereographic projection S' - {(0,1)} → R of example 2* is given by the formula (x,y) → 2x/(1-y) and derive the formula for the projection S² - {(0,1)} → R².

6. Show that the cube [0,1]³ = {(x,y,z) ∈ R³ : 0 ≤ x, y, z ≤ 1 is homeomorphic with the solid 3-sphere:
{(x,y,z) ∈ R³: x² + y² + z² ≤ 1}.

Example 1*
Let L be the real line together with an extra point which we'll call 0' and think of as an extra 0. We make this into a topological space in the following way. For every subset of R that does not contain 0, there is corresponding subset of L, and we define this subset of L to be open if the corresponding subset of R is open. For each subset of R that contains 0, there are corresponding subsets of L: One which contains 0, one which contains 0' and one which contains both. We define all three to be open if the original subset of R is open. The open sets on L that this gives do actually form a topology, i.e., they satisfy the axioms T1 - T4, as you can check.
This construction gives a curious space called the real line with a double point at 0.

Example 2*
Since Sⁿ is connected for n > 0, this shows that RPⁿ is connected for n > 0.
The converse is not true: X / ~ may be connected even though X is disconnected, as the following example shows.
The set {1,2}, with the indiscrete topology, is not Hausdorff. For if we take x = 1 and y = 2, then the only open set containing x is the whole set, which also contains y. Thus it is not possible to find two non-overlapping open sets each containing only one of the points.

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