Topology on quotient spaces: punctured lines I am studying topology and I have some doubt about the following. I show what I have tried. Any help would be really appreciated!
Let X be the quotient space of the Euclidean Plane (with usual topology) w.r.t. the equivalence generated by "x is equivalent to y IFF they are both not the origin and they lie on the same line through the origin".
Then the members of the set X are the origin and all the lines through it (but without it). Correct?
1) Describe the open sets and the nbhd filter on each point.
No singleton in X is open. Unions of puctured lines are open if they are obtained by continuously rotating one such line by a certain nonzero angle.
A union of punctured lines and the origin is open if the same condition as before is satisfied. These 2 kinds of unions are a basis for the topology.
In particular the punctured plane is open.
The origin has as nbhds the portions of the planes that contain the open sets of the second kind previously defined.
A punctured line has both kind of open sets in its nhbd filter.
2) Is X separable?
The punctured lines with rational angular coefficient, together with the origin, are a dense coiuntable subset. (X is separable.)
It seems to me there is no countable basis. Is that correct?
However, it is C1.
X is T1 because singletons are closed. It is not T2 beacuse the origin and one punctured line cannot be separated by open sets. So it is not regular, hence not completely regula, hence not pseudometrizable, hence not metrizable.
3) Open sets are also closed. So X is not connected (and not path-conected).
What about locally (path-)connectedness?
4) X is not compact, but locally compact.
Finally, show that X is in set-theoretical bijection with S = the unit circle plus the origin. Compare the topology of the quotient space above with the topology induced by the plane on S. I am not sure how to procede here.
 A: Let $p : \mathbb{R}^2 \to X$ denote the quotient map. Set $\ast = p(0)$, $\mathbb{R}^2_0 = \mathbb{R}^2 \backslash \lbrace 0 \rbrace$ and $X_\ast = X \backslash \lbrace \ast \rbrace$. We have $p^{-1}(\ast) = \lbrace 0 \rbrace$ so that $p$ restricts to a quotient map $q : \mathbb{R}^2_0 \to X_\ast$.
$X_\ast$ is an open subspace of $X$; $\ast$ has only one open neighborhood - the whole space $X$ (for any open neighborhood $U$ of $\ast$ we get an open neighborhood $p^{-1}(U)$ of $0$ in $\mathbb{R}^2$ which must meet all punctured lines so that $U = p(p^{-1}(U)) = X$).
The space $X_\ast$ is nothing else than the real projective line $\mathbb{R}P^1$ (see e.g. https://en.wikipedia.org/wiki/Projective_space).
This is known to be homeomorphic to the circle $S^1$.
For the sake of completeness let us explicitly construct a homeomorphism $h : X_\ast \to S^1$. We identify $\mathbb{R}^2$ with $\mathbb{C}$ in order to make use of the complex multiplication. We have $S^1 = \lbrace z \in \mathbb{C} \mid \lvert z \rvert = 1 \rbrace$. Define $s : \mathbb{C} \backslash \lbrace 0 \rbrace \to S^1, s(z) = z^2/\lvert z^2 \rvert$. This is a continuous surjective map. Let $l_u$ denote the punctured line going through $u \in S^1$, i.e. $l_u = \lbrace t u \mid t \in \mathbb{R} \backslash \lbrace 0 \rbrace \rbrace$. We have $l_u = l_v$ iff $u= \pm v$. Clearly $s$ maps each $l_u$ to the same point of $S^1$ so that there exists a unique continuous map $h : X_\ast \to S^1$ such that $h \circ q = s$. Explicitly, $h(l_u) = u^2$. Since $s$ is surjective, so is $h$. It is also injective since $h(l_u) = h(l_v)$ iff $u^2 = v^2$ iff $u= \pm v$. Hence $h$ is a bijection and has an inverse $h^{-1} : S^1 \to X_\ast$. It remains to show that $h^{-1}$ is continuous. Let $w = e^{i\alpha} \in S^1$. Then $h^{-1}(w) = l_{e^{i\alpha/2}}$. Let $U$ be an open neighborhood of $h^{-1}(w)$ in $X_\ast$. Then $V = q^{-1}(U)$ is open in $\mathbb{C} \backslash \lbrace 0 \rbrace$ and $e^{i\alpha/2} \in V$. There exists $\epsilon > 0$ such that $e^{i\beta/2} \in V$ for $\lvert \beta - \alpha \rvert < \epsilon$. But $V$ is a union of punctured lines so that $l_{e^{i\beta/2}} \subset V$ for $\lvert \beta - \alpha \rvert < \epsilon$. This means $h^{-1}(e^{i\beta}) \in U$ for $\lvert \beta - \alpha \rvert < \epsilon$. But $W = \lbrace e^{i\beta} \mid \lvert \beta - \alpha \rvert < \epsilon \rbrace$ is an open neighborhood of $w$ in $S^1$ such that $h^{-1}(W) \subset U$.
Now, define a set $S^1_0 = S^1 \cup \lbrace 0 \rbrace$. $h$ extends to a bijection $H : X \to S^1_0$ and we give $S^1_0$ the topology such that $H$ becomes a homeomorphism. We can therefore consider $S^1_0$ instead of $X$. $S^1$ with its standard topology is an open subspace of $S^1_0$. $0$ has only one open neighborhood - the whole space $S^1_0$.
This implies that $S^1_0$ is separable and has a countable base. Open sets are in general not closed (an example is $S^1$). $S^1_0$ is not Hausdorff, but (quasi-)compact, i.e. each open cover has a finite subcover. It is also locally (quasi-)compact. Moreover, it is pathwise connected: For each $x \in S^1$ we have a path $u : [0,1] \to S^1_0, u(0) = 0, u(t) = x$ for $t > 0$.
The comparison of $S^1_0$ with its above toplogy and the topology inherited from $\mathbb{R}^2$ is now obvious.
