Let $X$ be the disjoint union of copies of a subspace of $\mathbb{R}$. Is the quotient of $X$ by identifying accumulation points first countable?

Question

Sorry the title isn't super clear, I ran out of characters.

From Willard's General Topology, 16A.5:

For each $n \in \mathbb{N}$, let $X_n$ be a copy of the subspace $\{0\} \cup \{\frac{1}{m}: m = 1, 2, \ldots \}$ of $\mathbb{R}$. Let $X$ be the disjoint union of the $X_n$. Is the quotient $Y$ of $X$ obtained by identifying all accumulation points of $X$ first countable?

I'm really having a hard time with this question. I'm struggling more generally with both the notions of the disjoint union topology, and the quotient topology, so this is really throwing me for a loop.

The relevant information that I know is:

1. If $X$ is a topological space and $a \in X$, then we say $a$ is an accumulation point of a set $A$ iff each nbhood of $a$ meets $A$ in some point other than $a$,
2. The set $X_n'$ of all accumulation points of $X_n$ (defined above) is $X_n' = \{0\}$.

But I'm not really sure what $X$, the disjoint union of these $X_n$, "looks" like. Is the set of accumulation points of $X$ the union of the accumulation points of the $X_n$? I'm really just not sure how to proceed.

Any help is appreciated.

Answer 1

$X$ is homeomorphic to $\Bbb Z^+\times X_1$ with the product topology, where $\Bbb Z^+$ has the discrete topology. $\Bbb Z^+\times X_1$, and therefore also $X$, looks something like the diagram below: the points, symbolized by labelled bullets, are ordered pairs $\left\langle k,\frac1n\right\rangle$ for $k\in\Bbb Z^+$ and $\frac1n\in X_1$.

$$\begin{array}{rrr} \langle 1,1\rangle\;\bullet&\langle 1,1\rangle\;\bullet&\langle 1,1\rangle\;\bullet&\langle 1,1\rangle\;\bullet&\cdots\\ \left\langle 1,\frac12\right\rangle\bullet&\left\langle 1,\frac12\right\rangle\bullet&\left\langle 1,\frac12\right\rangle\bullet&\left\langle 1,\frac12\right\rangle\bullet&\cdots\\ \left\langle 1,\frac13\right\rangle\bullet&\left\langle 1,\frac13\right\rangle\bullet&\left\langle 1,\frac13\right\rangle\bullet&\left\langle 1,\frac13\right\rangle\bullet&\cdots\\ \left\langle 1,\frac14\right\rangle\bullet&\left\langle 1,\frac14\right\rangle\bullet&\left\langle 1,\frac14\right\rangle\bullet&\left\langle 1,\frac14\right\rangle\bullet&\cdots\\ \vdots\,&\vdots\,&\vdots\,&\vdots\,\\ \langle 0,0\rangle\;\bullet&\langle 0,0\rangle\;\bullet&\langle 0,0\rangle\;\bullet&\langle 0,0\rangle\;\bullet&\cdots \end{array}$$

To get $Y$, just glue all of the $0$ points together in a single point: you end up with a sort of hedgehog, with countably infinitely many spikes — the sets $X_n\setminus\{0\}$, which are columns in the picture above — sticking out of the central point where all of the $0$s are glued together. Call that central point $p$.

HINT: You should be able to check easily that points of $Y\setminus\{p\}$ are isolated, so $Y$ is first countable at each of those points.

If $\sigma=\langle k_n:n\in\Bbb Z^+\rangle$ is any sequence of positive integers, let $$B(\sigma)=\{p\}\cup\bigcup_{n\in\Bbb Z^+}\left\{x\in X_n:x\le\frac1{k_n}\right\}\,.$$

Check that each of these sets $B(\sigma)$ is an open nbhd of $p$ and that in fact they form a local base at $p$. Then show that if $\mathscr{V}$ is any countable family of open nbhds of $p$, there is a $B(\sigma)$ that does not contain any member of $\mathscr{V}$, so $\mathscr{V}$ cannot be a local base at $p$. This shows that $Y$ is not first countable at $p$.

Answer 2

This is a very similar question as Some topological properties of "countable lines with one origin".

Let $L = \{0\} \cup \{\frac{1}{m}: m \in \mathbb N \}$. Then we may take $X = L \times \mathbb N$, where $\mathbb N$ has the discrete topology (which is the subspace topology inherited from $\mathbb R \supset \mathbb N$). The set of accumulation points of $L$ is $L' = \{0\}$. Hence the set of accumulation points of $X$ is $X' = L' \times \mathbb N = \{0\} \times \mathbb N$. Certainly no $(x,n) \in X$ with $x \ne 0$ is an accumulation point of $X$ because $\{(x,n)\} = \{x\} \times \{n\}$ is open in $X$. Now consider $(0,n) \in X$ and an open neigborhood $U$ of $(0,n)$ in $X$. There exist open neigborhoods $U_n$ of $0$ in $X$ and $V_n$ of $n$ in $\mathbb N$ such that $U_n \times V_n \subset U$. The set $U_n$ contains some $p \in L$ with $p \ne 0$. Hence $(p,n) \in U$ and $(p,n) \ne (0,n)$.

This means that $Y = X/X'$. Let $p : X \to Y$ denote the quotient map and $*$ the common equivalence class of the points $(0,n)$. For $M \in \mathbb N$ let $L_M = \{0\} \cup \{\frac{1}{m}: m \ge M \}$ which is an open neigborhood of $0$ in $L$.

Let $\{U_k\}$ be any countable family of open neigborhoods of $*$ in $Y$. Then $p^{-1}(U_k)$ is an open neigborhood of $(0,n)$ in $X$, thus there exists $M(n)$ such that $L_{M(n)} \times \{n\} \subset p^{-1}(U_k)$. The set $V = p(\bigcup_{n= 1}^\infty L_{2M(n))} \times \{n\})$ is an open neighborhood of $*$ in $Y$ because $p^{-1}(V) = \bigcup_{n= 1}^\infty L_{2M(n))} \times \{n\}$ which is open in $X$. For each $n$ we have $p^{-1}(U_n) \not\subset p^{-1}(V) = \bigcup_{n= 1}^\infty L_{2M(n))} \times \{n\}$ because $(0,M(n)) \in p^{-1}(U_n)$, but $(0,M(n))\notin \bigcup_{n= 1}^\infty L_{2M(n))} \times \{n\}$. This means $U_n \not\subset V$, thus $\{U_k\}$ cannot be a neighborhood base of $*$.

Answer 3

The quotient topology on $Y$ is defined as the strongest topology such that the quotient map $q:X\to Y$ is continuous. Let $T_X$ be the topology on $X.$

Let $T_Y=\{S\subseteq Y: q^{-1}S\in T_X\}.$ Then $T_Y$ is a topology on Y, merely because $q$ is surjective and because $T_X$ is a topology on $X$.

(1). If $T$ is a topology on $Y$ with $T\supsetneqq T_Y$ then there exists $S'\in T$ \ $T_Y,$ which by def'n of $T_Y$ implies $q^{-1}S'\not \in T_X,$ so $q$ is not continuous with the topology $T$ on $Y.$

So $T_Y$ is the quotient topology on $Y.$

Let $E$ be an equivalence relation on $X.$ The quotient map $q$ induced by $E$ is $q(x)=[x]_E=\{x'\in X:x'Ex\}.$ This can be notationally and conceptually inconvenient. When $\phi\ne C\subset X$ and $[x]_E=\{x\}$ for $x\in X$ \ $C$ while $[x]_E=C$ for $x\in C,$ we can take some (any) $p\not \in X,$ and let $q(x)=x$ for $x\in X$ \ $C$ while $q(x)=p$ for $x\in C.$ The space $Y=\{p\}\cup (X \setminus C)$ with the quotient topology is called the identification of C to a point.

(2). For some set $I$, suppose $T_i$ is a topology on $X_i$ for each $i\in I.$ The topological sum (topological disjoint union) is $(X, T)$ where $X=\cup_{i\in I} (X_i\times \{i\})$ and $T$ is the topology on $X$ generated by the base (basis) $\cup_{i\in I}\{t\times \{i\}:t\in T_i\}.$ Note that each $X_i\times \{i\}$ is an open-and-closed subspace of $X$ and its subspace topology is $\{t\times \{i\}:t\in T_i\},$ so it is homeomorphic to the $T_i$ topology on $X_i.$ In particular, if $x=(x_i,i)\in X$ then $x$ is an isolated point of $X$ iff $x_i$ is an isolated point of $X_i.$