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Let the countable lines with one origin to be a quotient space $CL = ([0, \infty) \times \mathbb N) / \sim$, where $[0, \infty) \times \mathbb N$ has a subspace topology of $\mathbb R^2$ and $0 \times n \sim 0 \times m$ for any $n, m \in \mathbb N$.

Then, is $CL$ Hausdorff, first countable, and/or locally compact?

My attempts:

Hausdorff:

Take for any $x$ and $y$, if both are not $[0 \times n]$, take a disjoint neighborhoods of their correspondents in the original space, and pass it to the quotient space. Otherwise, if $x$ is $[0 \times n]$ and $y = [a \times b]$ is not, I can take something like $\pi([0, \frac{y}{2}) \times b)$ and $\pi((\frac{y}{2}, 2 y) \times b)$, although I am not completely sure if they'd be open.

First countable:

Let $\pi : [0, \infty) \times \mathbb N \to CL$ be a quotient map. If there exists a neighborhood basis $\{ U_n \}$ around $[0 \times 1] \in CL$, I believe I can take something like $U = \bigcup \limits_{n = 1}^\infty \pi([0, \frac{r_n}{2}) \times n)$ where $[0, r_n) \times n \subset \pi^{-1} (U_n) \cap [0, \infty) \times n$, so that it is not properly contained in any of the neighborhood basis, but not sure if it is open again :(.

Locally compact:

Not sure...

Trash Failure
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  • You can consider $\pi([0,\frac y2)\times\Bbb N)$ instead of $\pi([0,\frac y2)\times b)$. – Greg Martin Oct 30 '20 at 08:10
  • Does it make any difference? – Trash Failure Oct 30 '20 at 08:13
  • One of them is open, the other is not. – Greg Martin Oct 30 '20 at 08:14
  • Why is it? I do not see why would the former be necessarily open, and other is not. – Trash Failure Oct 30 '20 at 08:16
  • Are you assuming $\pi$ is an open map? Even if so, I think both should be open. – Trash Failure Oct 30 '20 at 08:22
  • This is a great opportunity to revisit the definition of the quotient topology; it's defined precisely to be the (simplest) topology that makes $\pi$ an open map. As another way to guide your intuition, replace $\Bbb N$ by a two-element set ${+,-}$, so that the quotient space is homeomorphic to $\Bbb R$ itself; is $[0,\frac y2) \times {+} \sim [0,\frac y2)$ open in that quotient? – Greg Martin Oct 30 '20 at 15:50
  • Your argument showing that $CL$ is not first countable at the origin is correct, if you’re using $\subset$ for proper inclusion (i.e., $\subsetneqq$). – Brian M. Scott Oct 30 '20 at 21:17
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    For local compactness observe that if $U$ is any open nbhd of the origin, there are points $\pi(\langle x_n,n\rangle)\in U$ for $n\in\Bbb N$ with $x_n\ne 0$. Show that ${\pi(\langle x_n,n\rangle):n\in\Bbb N}$ is a closed, discrete subset of $\operatorname{cl}U$, which therefore cannot be compact. – Brian M. Scott Oct 30 '20 at 21:22

1 Answers1

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Let $p : [0,\infty) \times \mathbb N \to CL$ denote the quotient map and $* \in CL$ denote the common equivalence class of the points $(0,n)$ with $n \in \mathbb N$. Clearly $p^{-1}(*) = \{0\} \times \mathbb N$ which is closed in $[0,\infty) \times \mathbb N$. Thus $\{*\}$ is closed in $CL$ and $CL' = CL \setminus \{*\}$ is open in $CL$.

  1. $CL$ is Hausdorff.

The restriction $p' : (0,\infty) \times \mathbb N \to CL'$ of $p$ is a continuous bijection and it is easy to see that it is an open map. Hence $p'$ is a homeomrphism. This shows that any two distinct $x,y \in CL'$ have disjoint open neigborhoods in $CL$. Now let $x = *$ and $y \in CL'$. We can write $y = p(t,n)$ for some $t > 0$ and $n \in \mathbb N$. Then $U = p((t/2,\infty) \times \{n\})$ is an open neigborhood of $y$ and $V = p([0,t/2) \times \mathbb N)$ is an open neigborhood of $*$. We have $U \cap V =\emptyset$.

  1. $CL$ is not first countable.

Of course each $x \in CL'$ has a countable neigborhood base. The exceptional case is $*$. Let $\{U_k\}$ be any countable family of open neigborhoods of $*$ in $CL$. Then $p^{-1}(U_k) = \bigcup_{n= 1}^\infty U_k^n \times \{n\}$ with open neighborhoods $U_k^n$ of $0$ in $[0,\infty)$. There are $t_k^n > 0$ such that $[0,t_k^n) \subset U_k^n$. Define $s_n = \dfrac{t_n^n}{2}$ (this is a typical diagonal construction). Then $V = \bigcup_{n= 1}^\infty [0,s_n) \times \{n\}$ is open in $[0,\infty) \times \mathbb N$. We have $p^{-1}(p(V)) = V$, thus $U = p(V)$ is an open neigborhood of $*$. For each $n$ we have $p^{-1}(U_n) \not\subset p^{-1}(U)$ because $[0,t_n^n) \times \{n\} \not\subset [0,s_n) \times \{n\}$. This means $U_n \not\subset U$, thus $\{U_k\}$ cannot be a neighborhood base of $*$.

  1. $CL$ is not locally compact.

Of course each $x \in CL'$ has a compact neigborhood. The exceptional case is again $*$. Let $U$ be any open neigborhood of $*$. We shall show that $\overline U$ is not compact. We have $p^{-1}(U) = \bigcup_{n= 1}^\infty U_n \times \{n\}$ with open neighborhoods $U_n$ of $0$ in $[0,\infty)$. There are $t_n > 0$ such that $[0,t_n] \subset U_n$. The set $V = \bigcup_{n= 1}^\infty [0,t_n) \times \{n\}$ is open in $[0,\infty) \times \mathbb N$. We have $p^{-1}(p(V)) = V$, thus $W = p(V)$ is an open neigborhood of $*$. The sets $W_n = p((\frac{t_n}{2},\infty) \times \{n\})$ are open in $CL$ and we have $W \cup \bigcup_{n=1}^\infty W_n = CL \supset \overline U$. For each $m$ we have $p(t_m,m) \notin W \cup \bigcup_{n \ne m} W_n$, and since $(t_m,m) \in U_m \times \{m\} \subset p^{-1}(U)$, i.e. $p(t_m,m) \in U \subset \overline U$, we see that $\{W\} \cup \{W_n\}$ does not contain a finite subcover of $\overline U$.

Paul Frost
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