# Determine if these spaces are connected, Hausdorff, or compact.

Let $$X = [0,1]/(0,1)$$ and let $$\pi: [0,1] \rightarrow X$$ be the quotient map. Answer the following questions, proving your assertions:

a) Is $$X$$ contractible?

We need $$s:X \rightarrow \{pt\}$$ and $$t: \{pt\} \rightarrow X$$ with $$s \circ t \simeq id_{\{pt\}}$$ and $$t \circ s \simeq id_{X}$$.

$$s$$ is the obvious map. Define $$t$$ with $$pt \mapsto 0$$. We have $$s \circ t = id_{\{pt\}}$$. Now define $$H(x,t) = (t \circ s)(1-t) + id_{X}(t)$$, which is a homotopy from $$t \circ s$$ to $$id_{X}$$. So Yes, $$X$$ is contractible.

b) Are $$\pi(0)$$ and $$\pi(1/2)$$ dense in $$X$$?

$$\{0\}$$ is closed in $$X$$ because $$\pi^{-1}(\{0\})$$ is closed in $$[0,1]$$. So $$\overline{\{0\}} \not = X \implies \pi(0)$$ is not dense in $$X$$.

Since $$\overline{\pi(1/2)}=\overline{(0,1)} = [0,1]$$, $$\pi(1/2)$$ must be dense in $$X$$.

c) Consider the complement of $$\pi(0)$$ in $$X$$. Is it Hausdorff? Is it connected?

No it's not Hausdorff. For $$[1/2]$$ and $$1$$, we need to find $$U$$ and $$V$$ open and disjoint in $$X$$ with $$[1/2] \in U$$ and $$1 \in V$$. But we need $$V \subseteq X - (0,1)$$. So $$V = \{1\}$$, which is closed (since $$\pi(\{1\})$$ is not open in $$[0,1]$$).

It is connected. Suppose we have $$X=A \cup B$$ for $$A,B$$ open, nonempty, and disjoint in $$X$$. Since there are only two elements, the only possibilities for $$A$$ and $$B$$ are $$(0,1)$$ and $$\{1\}$$. But $$\pi(\{1\})$$ is not open so, so $$\{1\}$$ cannot be open in $$X$$. By the definition of the subspace topology, it is not open in $$X - \pi(0)$$ either.

d) Consider the complement of $$\pi([1/2])$$ in $$X$$. Is it Hausdorff? Is it connected?

Yes it is Hausdorff. Since $$X - \{0\}$$ is open in $$X$$, $$(X -\{0\}) \cap (X - \pi([1/2]) = \{1\}$$ is open in the subspace topology. Similarly for $$\{0\}$$. Hence the subspace is Hausdorff.

It's not connected since $$X - \pi([1/2]) = \{0\} \cup \{1\}$$.

As I see, your answers are correct. There are minor issues with your definition of $H$. One thing is that $t$ stands there both for name of a function and a variable. Another is that you pass $t ∈ [0, 1]$ to $id_X$ but $X$ is not $[0, 1]$, there should be composition with the projection. But the idea is clear: you take the standard homotopy for contraction of $[0, 1]$ and show that it factors through $π$.
For other questions you may also forget $[0, 1]$ and see that $X$ is just three point space with a certain topology which can be easily visualized so the answers are clear.