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I want to solve the following question:

Prove that the union of $W$ and the unit circle $S^1$ is connected in the subspace topology of $\mathbb{R^2}$ where $W=\{(x, y) \in \mathbb{R^2} | x=(1-e^{-t})cost, y=(1-e^{-t})sint, t \geq0\}$

I know that a connected topological space, $X$ is connected if it does not split into open disjoint non-empty subsets.

Also, for a space $(Y, \tau)$ the subspace topology on $X \subset Y$ is $\tau|_X=\{U \cap X : U \in \tau\}$.

How can the union be connected, since you are taking the union of non-empty subsets?

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  • $\begingroup$ 'How can the union be connected, since you are taking the union of non-empty subsets?' $(0,2) \cup (1,3)=(0,3)$ which is connected. Don't forget the disjoint part of the definition! $\endgroup$ – noctusraid Apr 15 '16 at 11:40
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$W$ is connected (since it is a path). Note that $\overline{W}$ is connected, since it is the closure of a connected set. But $\overline{W}=W \cup S^1$.

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  • $\begingroup$ How is $(1,0)$ in $W$? I can only see that it should be an accumulation point of $W$. Isn't $|(1-e^{-t}) cos t| \leq |1-e^{-t}| < 1$ for $t\geq 0$? $\endgroup$ – sqtrat Apr 15 '16 at 12:57
  • $\begingroup$ @sqtrat Brain malfunctioning and assuming that $e^0=0$. $\endgroup$ – Aloizio Macedo♦ Apr 15 '16 at 13:44
  • $\begingroup$ @AloizioMacedo thanks. Can you elaborate as to why the closure is this union? Does it have something to do with limit points? $\endgroup$ – amiz9 Apr 18 '16 at 16:40
  • $\begingroup$ @AloizioMacedo Thank you for the reply. I was wondering how you knew that $\bar{W}=W\cup S^1$ $\endgroup$ – thinker May 2 '16 at 17:18
  • $\begingroup$ @thinker If you draw a figure, you can see that $W$ is spiralling towards $S^1$. If you want to prove it, just prove the following things: - First, any point in the circle is a limit point (this can be done by using the fact that $e^{-t} \to 0$ and the spiral keeps doing rotations). Secondly, any point out of the unit disk is not a limit point (this is obvious). Third, any point inside the unit disk which is not in the path is not a limit point. This follows from the fact that the spiral keeps going "outwards". Hence, it suffices to separate the point from a compact part of the path. $\endgroup$ – Aloizio Macedo♦ May 7 '16 at 6:54

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