Lee Problem 1.1

Let \(X^{\pm} = \{(x, \pm 1) \ | \ x \in \mathbb{R}\}\). Clearly, \(X = X^{+} \cup X^{-}\). Let \(X/\sim\) be the quotient space, which is topologized by taking the set of all \(U\) such that \(\pi^{-1}(U)\) is open in \(X\) as a subspace of \(\mathbb{R}^2\) (\(\pi\) is the quotient map).

For each equivalence class in \(X/\sim\) containing \((x, \pm 1)\), identify this class with \(x \in \mathbb{R}\). We let \(\mathcal{B}\) be the collection of all open balls centred at rational \(x\) with rational radius, not containing \(0\) (so the notion of an open ball is well-defined under our identification). Clearly, the inverse image under \(\pi^{-1}\) for any of these sets is open in \(X\).

Let \(\mathcal{B}'\) be the collection of all open balls of rational radius centred at \((0, 1)\) and \((0, -1)\), individually. Note that the inverse image under \(\pi^{-1}\) for these sets will be open as well:

\[\pi^{-1}(B_r((0, \pm 1))) = (-r, r) \times \{\pm 1\} \cup \left( (-r, 0) \cup (0, r) \right) \times \{\mp 1\}\]

Of course, \(\mathcal{B} \cup \mathcal{B}'\) is countable. It is also a basis for the topology: each element of the basis is open in the quotient topology as we have shown, and in addition, given a generic element of the topology, \(U\) with \(\pi^{-1}(U)\) open in \(X\), note that \(\pi^{-1}(U)\) can be written as a union of rational-radius balls centred at points along the two copies of \(\mathbb{R}\) not containing \(0\), as well as rational-radius balls centred at \(0\). \(\pi\) clearly sends these elements to elements of \(\mathcal{B} \cup \mathcal{B}'\), so \(\pi(\pi^{-1}(U)) = U\) (as \(\pi\) is surjective) is a union of elements of the countable set \(\mathcal{B} \cup \mathcal{B}'\).

It follows immediately that \(\mathcal{B} \cup \mathcal{B}'\) is a countable basis for the quotient topology.

This space is also clearly locally Euclidean: the charts sending the two open sets \(U_{\pm} = \{ x \neq 0 \} \cup (0, \pm 1)\) to their projection onto the first coordinate are homeomorphisms \(\varphi_{\pm} U_{\pm} \rightarrow \mathbb{R}\).

Finally, note that the space is not Hausdorff. Suppose \(V_{\pm}\) are open sets about \((0, \pm 1)\), so \(\pi^{-1}(V_{\pm})\) are open in \(X\). Then the \(\pi^{-1}(V_{\pm})\) must each contain some interval \((-\varepsilon, \varepsilon) \cup \{ \pm 1\}\) respectively: but then \(V_{+} \cap V_{-} \neq \emptyset\).