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Does anybody know the Alexandroff's double segment space? References would be very welcome.

I will try to describe it here:

Alexandroff's double segment space: Suppose $X = C_1 \cup C_2$, where $C_1$ and $C_2$ are the two segments in $\mathbb R^2$ given by $C_i=\{(x,y): y=i, 0\le x \le 1 \}$, $i=1,2$. The neighborhoods of the points in $X$ are given as follows: If $x \in C_2$, then $B(x)=\{\{x\}\}$; and if $x=(a,1)\in C_1$, then $B(x)=\{U_k(x)\}^\omega_{k=1}$, where $U_k(x)=\{(a', b'): 0 \lt |a-a'| \lt \frac1k \} \cup \{x\}.$

If we identify all the points in $C_1$ to one point, say $c^*$, i.e., we get a new quotient topology of Alexandroff's double segment space. My textbook said the weight of nbhd base of $c^*$ is uncountable.

I don't know how to show it, I can't even understand the form of its nbhd base.

Could anybody help me? Thanks ahead :)

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Do you mean Alexandroff double circle? –  Temitope.A Jul 22 '12 at 13:42
    
@Temitope.A No. It's another topological space. –  Paul Jul 23 '12 at 0:05
    
@Nate Eldredge: Thanks for your useful advice:) –  Paul Jul 23 '12 at 0:06
    
@Nate Eldredge I have revised my question, and i hope the question could be reopen. –  Paul Jul 23 '12 at 0:48
2  
Thanks t.b. it is a nice job; I'm very appreciated your work. My text book which i am reading is a chinese book. It's not Engelkin's. Thank you again:) –  Paul Jul 23 '12 at 1:21

2 Answers 2

up vote 2 down vote accepted

Here’s a very rough sketch of $X$:

                         V     W
               --------(===)o(===)------------  

               --------(====x====)------------  
                       a    U    b

The top line represents $C_2$ and the bottom line $C_1$. The diagram shows a basic open nbhd of $\langle x,1\rangle\in C_1$: the nbhd consists of the open interval $U=\big(\langle a,1\rangle,\langle b,1\rangle\big)$ about $\langle x,1\rangle$ in $C_1$ together with the two open intervals $V=\big(\langle a,2\rangle,\langle x,2\rangle\big)$ and $W=\big(\langle x,2\rangle,\langle b,2\rangle\big)$ in $C_2$. Points of $C_2$, on the other hand, are isolated.

Now let $Y$ be the result of identifying $C_1$ to a point $c^*$, and let $q:X\to Y$ be the quotient map, so that $q(x)=c^*$ for $x\in C_1$ and $q(x)=x$ for $x\in C_2$. Points of $C_2$ are still isolated. Suppose that $c^*\in H\subseteq Y$; then $H$ is open iff $q^{-1}[H]=C_1\cup\big(H\setminus\{c^*\}\big)$ is open in $X$. Thus, we want to know the answer to the following question:

For what sets $A\subseteq[0,1]$ is $C_1\cup\big( A\times\{2\}\big)$ open in $X$?

Suppose that $C_1\cup\big( A\times\{2\}\big)$ is open. Then for each $x\in[0,1]$ there must be an open interval $(a,b)$ such that $x\in(a,b)$ and $$\Big((a,x)\cup(x,b)\Big)\cap[0,1]\subseteq A\;,$$

to ensure that $C_1\cup\big( A\times\{2\}\big)$ contains the basic open nbhd $$\left(\Big((a,b)\times\{1\}\Big)\cup\Big(\big((a,x)\cup(x,b)\times\{2\}\Big)\right)\cap X$$ of $\langle x,1\rangle$.

This can be restated as follows: for each $x\in(0,1)$ there are $a,b\in(0,1)$ such that $x\in(a,b)$ and $A\supseteq(a,b)\setminus\{x\}$, and there are also $a,b\in(0,1)$ such that $A\supseteq(0,a)$ and $A\supseteq(b,1)$. In even simpler terms, each point of $(0,1)$ has an open nbhd in $[0,1]$ that contains at most one point of $[0,1]\setminus A$.

Added: To see this, suppose that $x\in(0,1)$. (The argument for $0$ and $1$ is very similar.) By the previous sentence there are $a,b\in(0,1)$ such that $x\in(a,b)$ and $A\supseteq(a,b)\setminus\{x\}$. If $x\in A$, then $A\supseteq(a,b)$, and $(a,b)$ is an open nbhd of $x$ in $[0,1]$ that is disjoint from $[0,1]\setminus A$. If $x\notin A$, then $(a,b)$ is still an open nbhd of $x$ in $[0,1]$, and the only point of $[0,1]\setminus A$ in $(a,b)$ is $x$ itself. Thus, in either case $(a,b)$ contains at most one point of $[0,1]\setminus A$.

But that just says that $[0,1]\setminus A$ is a closed, discrete set in $[0,1]$, which means that $[0,1]\setminus A$ must be finite.

In other words, if $c^*\in H\subseteq Y$, then $H$ is open in $Y$ iff $H$ contains all but finitely many points of $C_2$. Let $\mathscr{B}$ be the collection of all such subsets of $Y$; it’s easy to show that if $\mathscr{C}$ is a countable subset of $\mathscr{B}$, there is a $B\in\mathscr{B}$ that does not contain any $C\in\mathscr{C}$, so $\mathscr{C}$ cannot be a local base at $c^*$. Thus, $Y$ is not first countable at $c^*$.

share|improve this answer
    
@Paul: The first inclusion is backwards, and the $1$ should be $a$. –  Brian M. Scott Aug 10 '12 at 7:38
    
@Brain Yes. It should be this:) –  Paul Aug 10 '12 at 7:41
    
@Brain Why does each point of $A$ have an open nbhd in $[0,1]$ that contains at most one point of $[0,1]\setminus A$? It is a little difficult for me? Could you explain more? –  Paul Aug 10 '12 at 7:55
    
@Paul: Are you okay with the sentence before that one? –  Brian M. Scott Aug 10 '12 at 8:10
    
@Brain It's no problem for that sentence. –  Paul Aug 10 '12 at 8:21

I presume you mean the Alexandroff duplicate of $\mathbb{R}$, right? The 'duplicate' construction is valid for any $T_1$-space and it is quite well-known.

You might find this article useful:

share|improve this answer
    
Thanks Vitalis for your answer and the link of this article. The article is very interesting:) –  Paul Jul 23 '12 at 0:09

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