Exercise 2.21

Question

Exercise 21:
    Let $A$ and $B$ be separated subsets of some $\mathbf R^k$, suppose $\mathbf a\in A$, $\mathbf b\in B$, and define $$\mathbf p(t)=(1-t)\mathbf a+t\mathbf b$$
    for $t\in\mathbf R$. Put $A_0=\mathbf p^{-1}(A)$, $B_0=\mathbf p^{-1}(B)$.
    \begin{itemize}
        \item[(a)] Prove that $A_0$ and $B_0$ are separated subsets of $\mathbf R$.
        \item[(b)] Prove that there exists $t_0\in(0,1)$ such that $\mathbf p(t_0)
otin A\cup B$.
        \item[(c)] Prove that every convex subset of $\mathbf R^k$ is connected.
    \end{itemize}

ghostofgarborg · Accepted Answer

\def\eps{\varepsilon}

(a) We note that when $\epsilon > 0$
    $$ \| \mathbf p (t \pm \epsilon) - \mathbf p(t) \| = \epsilon \| \mathbf b - \mathbf a \| $$
    and if $ | t - t^\prime | < \epsilon / \| \mathbf b - \mathbf a \| $, then 
    $ \| \mathbf p(t) - \mathbf p(t^\prime) \| < \epsilon $. Therefore, if $\mathbf p(t) = \mathbf y$, then
    $$ N_{\epsilon / \| \mathbf b - \mathbf a\|}(t) \subset \mathbf p^{-1}(N_\epsilon(\mathbf y)) $$
    Consequently, the inverse image under $\mathbf p$ of an open set is open, and the inverse image of a closed set
    is closed. Clearly $A_0$ and $B_0$ are disjoint and non-empty, so this allows us to conclude that
    $$ \bar A_0 \cap B_0 \subseteq \mathbf p^{-1}(\bar A) \cap \mathbf p^{-1}(B) = \mathbf p^{-1}(\bar A \cap B) = \emptyset $$
    And similarly when interchanging $A$ and $B$. This shows that $A_0$ and $B_0$ are separated.

(b) Restrict $\mathbf p$ to $[0,1]$, and define $A_0$ and $B_0$ correspondingly. The result in (a) still holds. Since $[0,1]$ is connected, we must have $ \mathbf  [0,1] \supsetneq A_0 \cup B_0$. Choose 
    $t_0 \in [0,1] \setminus A_0 \cup B_0$. Then $t_0 \neq 0$ and $t_0 \neq 1$, and $\mathbf p(t_0) \not \in A \cup B$.

(c) Let $S \subset \mathbb R^k$. 
    The above result shows that if $\mathbf p(t) \in S$ for all $\mathbf a, \mathbf b \in S$ and
    $t \in [0, 1]$, then $S$ is connected. This is precisely the condition that $S$ is convex.

Amit Awasthi · Answer

\begin{proof}[Proof of $(a)$]
      Assume not. Then, there is a $t$ such that $t \in A_0 \cap \overline{B}_0$. This means $\mathbf{p}(t) \in A \cap B$. But since $A$ and $B$ are separated subsets of $\mathbb{R}^k$, $\overline{A} \cap B = \emptyset = A \cap \overline{B}$, which contradicts that $\mathbf{p}(t) \in A \cap B$ since $\mathbf{p}(t)$ cannot be a member of the empty set.
    \end{proof}
    \begin{lemma}
      If a set $X$ is path connected then it is connected.
    \end{lemma}
    \begin{proof}[Proof of Lemma]
      The definition of path connectedness says there is a continuous map $f:[0,1] ightarrow X$ where $f(0) = x, f(1) = y$ for any given $x,y \in X$. Suppose $X$ is not connected, and let $X = A \cup B$ where $A, B$ are disjoint open sets. Then let $x \in A$ and $y \in B$. Since $X = A \cup B$, $[0,1] = f^{-1}(A) \cup f^{-1}(B)$ by the continuity of $f$, and this would be a separation of $[0,1]$ because they are disjoint in $[0,1]$ and are open because the inverse image of an open set is open by Theorem 4.8, which is a contradiction since $[0,1]$ is connected.
    \end{proof}
    \begin{proof}[Proof of $(b)$]
      Assume there is no such point $t_0$. This would mean the function $\mathbf{p}(t)$ would be a continuous map where $\mathbf{p}(1) = a$ and $\mathbf{p}(0) = b$, and thus $A$ and $B$ are path connected. However, by our Lemma above above path connectedness implies connectedness, which contradicts that $A$ and $B$ are separated, so $t_0$ must exist.
    \end{proof}
    \begin{proof}[Proof of $(c)$]
      Let $X$ be a convex subset of $\mathbb{R}^k$, which means $\lambda \mathbf{x} + (1 - \lambda) \mathbf{y} \in X$ whenever $\mathbf{x} \in X, \mathbf{y} \in Y$, and $0 < \lambda < 1$. Assume it is not connected, which means $X = A \cup B$ where $A$ and $B$ are separated. Then, we can construct a function $\mathbf{p}(t) = \lambda \mathbf{x} + (1 - \lambda) \mathbf{y}$, where $\mathbf{x} \in A$ and $\mathbf{y} \in B$. However, from $(b)$, we know that there is some $\lambda \in (0,1)$ where $\mathbf{\lambda} 
otin A \cup B$, which contradicts the fact that $X$ is convex. So, $X$ is connected.
    \end{proof}

Exercise 2.21

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Exercise 2.21

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