Take the 2-minute tour ×
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.

This is a curious example I'm trying to understand.

For any positive integer $n$, let $A_n=\left\{\dfrac{1}{n}\right\}\times[0,1]$, and let $X=\bigcup_{n}A_n\cup\{(0,0)\}\cup\{(0,1)\}$. Viewing $X$ as a subset of the Euclidean plane, we can equip it with the relative topology.

Then why are $\{(0,0)\}$ and $\{(0,1)\}$ are components of $X$ and each closed and open subset contains both or neither of the points, but never just one or the other?

share|improve this question
This is very similar to what is called comb space: check this out en.wikipedia.org/wiki/Comb_space –  Bruno Stonek Dec 11 '11 at 19:28
Are you sure this is correct? In order to be a component of a space, the set has to be a subset of that space, and clearly neither $(0,0)$ nor $(0,1)$ belong to $X$, as you have defined it. –  Arthur Fischer Dec 11 '11 at 19:34
@ArthurFischer You're quite right, I meant to adjoin those points to $X$ as well. I've fixed that now. –  enkidu Dec 11 '11 at 19:43
add comment

1 Answer 1

up vote 2 down vote accepted

There are two things to address here:

1) Why are the singletons $\{(0,1)\}$ and $\{(0,0)\}$ components?

2) Why is there no closed and open subset of $X$ containing one point and not the other?

For the first question, recall that a component of $X$ is defined as a maximal connected subspace. So we need to show that the only connected subspace of $X$ containing, say, $(0,0)$, is the singleton set. That is, we need to show that a connected subspace $C \subset X$ containing $(0,0)$ must be the singleton. It should be readily apparent that $C$ cannot contain any point of $A_n$ for any $n$, because those points can be separated by a vertical line with irrational $x$-coordinate, and it immediately follows that it can't contain $(0,1)$ either.

As for the second question, any closed and open subset of $X$ must contain either all of or none of $A_n$, for each $n \in \mathbb{N}$. This is because $A_n$ is connected. So if $C \subset X$ is both open and closed, and contains $(0,0)$, then it contains $A_n$ for $n$ sufficiently large, ie, close to $(0,0)$ and $(0,1)$. Therefore $(0,1)$ is a limit point of $C$.

share|improve this answer
Thank you Dustan. –  enkidu Dec 11 '11 at 21:42
add comment

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.