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

A set S is called star-like if there exists a point $\alpha\in S$ such that the line segment connecting $\alpha$ and z is contained in S for all $z\in S$. Show that a star-like region is simply connected.

My answer

Show that $γ:γ(t)=tz+(1−t)α, t≥1$ is contained in the complement for any z in the complement

Let $\gamma$ represents the portion of the ray from $\alpha$ through z to $\infty$, starting at z. Thus, if z is in the complement of S, so is all of $\gamma$. For, if any $z_{1}\in \gamma$ belonged to S, so would the entire segment connecting $\alpha$ and $z_{1}$, including z.

Could anyone help to formalize the answer?

share|cite|improve this question
The assertion in your answer is false. The set $S=[-1,1]$ in $\Bbb R$ is star-like from the point $\alpha=0$, and $2$ is in the complement of $S$, but the ray from $0$ through $2$ includes the interval $[0,1]$, which is in $S$. To show that a star-like region $S$ is simply connected, you must show (1) that it is path-connected (which is easy), and (2) that every continuous map of the unit circle to $S$ is contractible to a point in $S$. – Brian M. Scott May 1 '12 at 3:12
I'm working on the complex plane – Breton May 1 '12 at 3:15
Doesn't matter: you can easily find similar examples there. – Brian M. Scott May 1 '12 at 3:15
I found the part that confuses the issue, to prove esque follow the suggestion of the book and what I proved but did not put it paste that I should prove, but already corrected – Breton May 1 '12 at 3:44
up vote 1 down vote accepted

If I recall right (see this), you only have to show that your star set is path connected and that all such paths are homotopic, i.e. that one path can be deformed continuously into the other.

Here's one way to do it : Let $z_1, z_2 \in S$. Then the path connecting $z_1$ to $\alpha$ and then $\alpha$ to $z_2$, is a continuous path between $z_1$ and $z_2$. Given any two paths, the first one between $z_1$ and $z_2$, and the other one between $z_3$ and $z_4$, just deform your first path continuously into a point at $\alpha$, and then "expand it" in a path from $z_3$ to $z_4$, which gives you the homotopy you need.

Hope that helps,

share|cite|improve this answer
If you wanna go formal on this, just parametrize the paths and parametrize the homotopy. – Patrick Da Silva May 1 '12 at 3:20

Here's a stronger algebraic-topological argument. We can show that a star-shaped set $S$ is not just simply connected, it is contractible. There is an $\alpha$ so for each $p\in S$ the line segment $\gamma_p(t) = tp + (1-t)\alpha$ has $\gamma_p\in S$ for all $t$. Now define $H(p,t):S\times [0,1]\to S$ by $(p,t)\mapsto \gamma_p(t)$. This is a strong deformation retract of $S$ onto $\{\alpha\}$, so $S$ has the homotopy type of a point.

For every topological space $X$, every map $f:X\to S$ is homotopic to the constant map $X\to\{\alpha\}$. The intuition here is to just "slide" the map to $\alpha$, more precisely via the homotopy $F(x,t) = H(f(x),t)$. So in particular we can see that any map $\phi:\mathbb{S}^1\to S$ can be contracted to the constant map $\phi:\mathbb{S}^1\to \{\alpha\}$, hence $\pi_1S = 1$.

share|cite|improve this answer

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.