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.

Show that the subspace of $\mathbb{R}^{3}$ that is the union of the spheres of radius $\frac{1}{n}$ and center $(\frac{1}{n},0,0)$ for $ n=1,2,3,...$ is simply-connected.

for showing it is simply-connected I must show that it is path connected and the its fundamental group is trivial.

I have little problem of what is happening for spheres when they are going to be near the point $(0,0,0)$,I mean that how should I investigate the properties at that point.

but my Idea for showing it is path connected is to take any two point in $X$,if they are in same sphere we are done,if they are in different spheres,we travel from one point to $(0,0,0)$ then from there to another point in another sphere.

for showing its fundamental group is trivial,my Idea is to swell any sphere to reach next bigger sphere and stick them together,when we do it for all of them, they become sphere with radius 1,which is contractible and have trivial fundamental group.because the spheres are infinite I have doubts.

please check my answer and make it right for me.thank you very much.

share|improve this question
It looks to me as all of them are contained in the biggest sphere, for $n=1$. Am I missing something? Did you maybe mean diameter rather than radius? –  Lubin Feb 22 at 17:57
I am sure that it is radius.and I don't think you missed any thing.they are inside each other where all of them exsect the origin. –  kpax Feb 22 at 18:33
Sorry, as happens all to often, I certainly was misreading the problem. –  Lubin Feb 22 at 20:55

2 Answers 2

up vote 1 down vote accepted

Since each sphere is simply connected, we can contract the piece of the loop that intersect with a particular sphere (that piece is a loop in that sphere with origin as the base). So for all the loop we just combine the contractions.

share|improve this answer

You're ok for this space to be path connected.

For showing it's simply connected, look at this link: http://en.wikipedia.org/wiki/Wedge_sum

Says you that whith Van Kampen theorem, the fundamental group is the free product of your fondamental groups, then 0. This construction is called wedge sum, or "bouquet", but, as has Daniel remarked, this is not exactly your case.

Remark: you say that a sphere is simply connected, what is absolutely true, and is contractible, which is absolutely false!!

share|improve this answer
It's not the wedge-sum, every neighbourhood of the origin contains almost all spheres completely, unlike for the wedge-sum. –  Daniel Fischer Feb 22 at 21:06
Yeah you're right, sorry. The wedge sum would be locally contractible, which is not the case here. But i think the Van Kampen argument still holds... –  Léo Feb 22 at 21:13
can you little explain more why it is not contractible?I was sure about that. –  kpax Feb 23 at 2:35
I confused,I must show that the fundamental group is trivial,the way that I show was right or wrong? –  kpax Feb 23 at 2:40
A sphere is not contractible, you know it, for example, cause the $H_2$ is not trivial! –  Léo Feb 24 at 21:28

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.