# Questions on the counterexample for “connectedness doesn't imply path-connectedness”

I am trying to understand the disproof/counterexample that 'a topological space that is connected may not be path connected'. Here is the explanation from "C Adams and R Franzosa - Introduction to Topology, Pure and Applied": I understood most of it but the following questions about the last 4 lines:

1- According to lemma 6.7. "Let $C$ and $D$ be subsets of a topological space $X$. Assume that $C$ is connected and $C\subset D$. Further assume that $U$ and $V$ form a separation of $D$ in $X$. Then either $C\subset U$ or $C\subset V$". But why in the 3rd line from end, $p(J)\subset U_n$ not $V_n$?

2- Why the statement "$p(J)\subset U_n\implies p(J)\subset S'\cap N'$" is true?

3- Why the statement "$J$ is a basis element for the topology on $[0, 1]$ implies that $p^{-1} (S^1)$ is open in $[0, 1]$", is true?

I truly appreciate any clear guidance.

1. Because, by the way we chose $y$, $y\in J$ and $p(y)\in S^1 \subset U_n$, so the lemma shows that $p(J)$ lies in the same connected component $p(y)$ does, which is $U_n$.
2. If $p(J)\in U_n$ for all $n$, then $p(J)\in \bigcap_{n=1}^\infty U_n$. This intersection is $S^1\cap N'$.
3. You have shows that every point $y\in p^{-1}(S^1)$ has an open neighborhood $J_y$ such that $y\in J_y \subset p^{-1}(S^1)$. So $p^{-1}(S^1)$ is a union of open sets (take the union over all $J_y$). Therefore it is open.
• Thank you very much. By the way, I forgot to ask also some question from the 3rd paragraph: Why $K=p([0,1])\subset S^1$ is equivalent to $p^{-1} (S^1)=[0,1]$, since $p^{-1} (K)=[0,1]$ is true? ($K\ne S^1$) And if I am right it makes a big difference to the whole text to be right at all, I think. – L.G. Jun 12 '15 at 13:46
• In other words of saying: It claims $p([0,1])\subset S^1$, meaning $p([0,1])$ may not 'cover' all of $S^1$. How is that all of $S^1$ has contribution in the domain of $p^{-1} (S^1)$? – L.G. Jun 13 '15 at 1:58
• @HIP13044b Look again at the definition of inverse image. If $A$ and $B$ are topological spaces, $f$ has domain $A$ and $f(A)\subset B$, then $f^{-1}(B)=A$, even if $A$ is properly contained in $B$. – Potato Jun 13 '15 at 2:03