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I'm having trouble proving that the subspace $X$ of $\mathbb{R}^2$ such that $X$ is the union of $[-1,1] \times \{ 0 \}$ and the line segments that join the points $(0,\frac{1}{n})$ with the point $(1,0)$ and the line segments from the points $(0,-\frac{1}{n})$ to $(-1,0)$ for all $n \in \mathbb{N}$ is not contractible to the point $(0,-1)$.

The only "progress" I've made in this problem is that I believe that the troublesome point with the contraction is the point $(0,0)$ and that I believe that $X$ is compact so that if there existed a contraction $H$, we would get that this function is uniformly continuous so any movement that a point in the neighborhood of $(0,0)$ made would force the point $(0,0)$ to move in the same direction because the uniform continuity and this could give a contradiction because the points in such neighborhood above the line $[-1,1] \times \{0\}$ move in the oposite direction as those below this line. However I'm not sure how can I write this in a more rigorous way or if the reasoning behind the idea is correct.

1 Answers1

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I think your approach works!

Let's establish some notation. Let $p_n = (0, \tfrac 1 n) \in X$ and let $q_n = (0, -\tfrac 1 n)$. Let $a = (1, 0)$ and let $b = (-1, 0)$. Let $r = (0, -1)$.

By the uniform continuity of $H$, there exists an $N \in \mathbb N$ such that $$ \left| H\left(p_N, s\right) - H\left(q_N, s\right)\right| < \tfrac 1 4 \ \ \ \ \ \ (\star)$$ for all $s \in [0, 1]$.

Next, I would use a connectedness argument to show that:

  • There exists an $s_1 \in [0, 1]$ such that $H(p_N, s_1) = a$, and such that for all $s \leq s_1$, $H(p_N, s)$ lies on the line segment between $p_N$ and $a$.
  • There exists an $s_2 \in [0, 1]$ such that $H(q_N, s_2) = b$, and such that for all $s \leq s_2$, $H(q_N, s)$ lies on the line segment between $q_N$ and $b$.

Now let $s' = \min(s_1, s_2)$. The inequality $(\star)$ is violated at time $s'$.

Let me fill in a few details about the "connectedness argument".

Let $g : [0, 1] \to X$ be the function $g(s) = H(p_N, s)$ and let $S$ be the set $$ S := \{ s \in [0, 1] : g(s) = a \} $$ Now observe:

  • $g([0, 1])$ is connected, since $g$ is continuous and $[0, 1]$ is connected.
  • $g([0, 1])$ contains the points $p_N$ and $r$, since $g(0) = p_N$ and $g(1) = r$.
  • $S$ is non-empty, since otherwise, $g([0, 1])$ would be a subset of $X \setminus \{ a \}$ containing the points $p_N$ and $r$, and any such subset is disconnected.
  • Therefore, it makes sense to define $s_1 := \inf (S)$.
  • $g(s_1) = a$, by the continuity of $g$.
  • $s_1 > 0$, since $g(0) \neq a$.
  • $g([0, s_1))$ is a subset of $X \setminus \{ a \}$.
  • $g([0, s_1))$ is connected, since $g$ is continuous and $[0, s_1)$ is connected.
  • $g([0, s_1))$ contains the point $p_N$, since $g(0) = p_N$.
  • Therefore, $g([0, s_1))$ must be contained within the line segment between $p_N$ and $a$. (In fact, $g([0, s_1))$ is the entirety of this line segment, though we don't actually use this.)
Kenny Wong
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