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"Let $A$ and $B$ be separated subsets of some $\mathbb{R}^k$, suppose $a \in A$, $b \in B$ and define $p(t) = (1-t)a + tb$, for $t \in \mathbb{R}$. Put $A_0 = p^{-1}(A), B_0 = p^{-1}(B)$

(a) Prove that $A_0$ and $B_0$ are separated subsets of $\mathbb{R}$.

(b) Prove that there exists $t_0 \in (0,1)$ such that $p(t_0) \notin A \cup B$.

(c) Prove that every convex subset of $\mathbb{R}^k$ is connected."

For the sake of completeness, I'll add that Rudin defines two sets in a metric space to be separated if each is disjoint from the other's closure.

I have managed to prove part (b), the way Rudin (probably) intended for it to be proved (a proof by contradiction with the aid of part (a)). However this was not how I first thought of the problem. My approach was to use a 'bisection argument' and the nested intervals theorem to locate the point $t_0$:

Recursively construct a sequence of intervals $I_n$, for $n = 0,1,2, ...$ by defining $I_0 = [0,1]$, and for every non-negative integer $k$, define $I_{k+1}$ using $I_k$ by bisecting $I_k$ and letting $I_{k+1}$ be the half which satisfies the following conditions

  1. $p(I_{k+1}) \nsubseteq A$
  2. $p(I_{k+1}) \nsubseteq B$

It is clear that such an $I_{k+1}$ exists. Thus we have constructed a sequence of compact nested intervals which means there is a point $t_0$ which is an element of every $I_n$.

Here is where I am stuck because intuition and the many pictures I drew tell me that this point is the one we seek to prove the question, however I am unsure of how to analytically prove that firstly $t_0$ is strictly an element of the open interval $(0,1)$ and that secondly $p(t_0) \notin A \cup B$.

I would appreciate any help in how to proceed from here :)

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Consider the following very bad scenario:

$$A_0=p^{-1}(A)=[0,1/2], \ \ B_0=p^{-1}(B)=[3/4,1].$$ Then your process may give $$ I_0=[0,1], $$ $$ I_1=[1/2,1], $$ $$ I_2=[1/2,3/4], \ \ I_3=[1/2,5/8], \ \ I_4=[1/2, 9/16], \cdots $$

Then the intersection of the nested intervals is $\{1/2\}$, and it belongs to $A_0$.

Because of this, the nested intervals method is not a good choice.

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  • $\begingroup$ If I may add one comment - there is one more possible problem. If both $A$, $B$ are singletons, you may choose the intervals in such a way that intersection of nested intervals will be $a$ or $b$. However, that is simple to fix - in the second step you have only to artificially choose the interval which image doesn't contain $a$ nor $b$. $\endgroup$ – Igor Sikora Jun 7 '18 at 22:00
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P = p([0,1]) is connected.
P $\cap$ A and P $\cap$ B are separated and not empty.
P $\cap$ (A $\cup$ B) is disconnected.
Thus P /= P $\cap$ (A $\cup$ B).
Whereupon be the beseached t in [0,1].

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