I have few of questions for the following proof in hatcher's book.

(1)Why f being continous imply that for each $s \in I$ has an open neighborhood $V_s$ in I mapped by f to some $A_{\alpha}$.

(2)Why we my take $V_s$ to be an interval whose closure is mapped to a single $A_{\alpha}$

In the second proof:

(3)I don't understand why we can express $S^n$ as $A_1,A_2$ such that each is homeomorphic to $R^n$, and $(A_1 \cap A_2) = S^{n - 1} \times \mathbb{R}$

(4)What is stopping us of doing the same thing for $n = 1$? enter image description here

  • $\begingroup$ When $n=1$ the double intersection is not path connected: it is a disjoint union of two segments. $\endgroup$ – Pedro Tamaroff Jun 15 '16 at 0:20
  • $\begingroup$ why did I get negative vote up ? $\endgroup$ – user329017 Jun 15 '16 at 0:25
  • 1
    $\begingroup$ @TheKEMO, please include your thinking in the body of the question as well. $\endgroup$ – Hmm. Jun 15 '16 at 0:34
  • $\begingroup$ I didn't know I should do that. Thank you for letting me know. $\endgroup$ – user329017 Jun 15 '16 at 0:49
  1. Since $f(s)$ lies in some $A_{\alpha}$ and $A_{\alpha}$ is open, there exists an open neighbourhood $V\subseteq A_{\alpha}$ of $f(s)$. Now, as $f$ is continuous, there exists an open neighbourhood $U \subset I$ of $s$ such that $f(U)\subseteq V$.

  2. We can take a $V$ such that $\overline V \subset A_{\alpha}$. Then, observe that $f(\overline U) \subseteq \overline{f(U)}\subseteq \overline V$.

  3. Think what happens when $n=2$. Take $A_1=\mathbb{S^2}\setminus\text{N}$, and $A_2=\mathbb{S^2}\setminus\text{S}$, where $N$ and $S$ denote the north pole and the south pole of $\mathbb{S^2}$ respectively. Then, clearly $A_1 \cap A_2$ is the equator, which is just $\mathbb{S^1}$.

  4. In this case, $A_1 \cap A_2$ is not path-connected.


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