Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :)

share|cite|improve this question
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
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
up vote 2 down vote accepted

Here’s a very rough sketch of $X$:

                         V     W

                       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|cite|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|cite|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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.