Rudin exercise 2.21 Clarification about convex sets being connected

Question

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

Prove $A_0$ and $B_0$ are separated subsets of $\mathbb{R}^1$

There was a similar question Baby Rudin's exercise 2.21, but the author's question in the comments was answered, and my question is similar to that question.

I'm having trouble seeing why $t \in A_0 \iff p(t) \in A$ is true.

What $A_0$ mean? Does it mean the set of all t that satisfies $p(t) \in A$ for ANY point $a \in A, b \in B$?

If it is for ALL $a \in A$, isn't $A_0$ simply 0, since then, $p(t) = (1-t)a + tb = a + 0*b = A$?

So regardless of what subsets $A$ and $B$ are, $A_0$ would always be ${0}$ and $B_0$ would always be ${1}$? Surely, this is not what the author intended?

But then again, if we interpret this so that there exists a point $a \in A$, $b \in B$, so that $p(t) \in A$, then $p(t)$ would really be dependent on the values of $a$ and $b$, since whether for a fixed $t$, $p(t) \in A$ would work for certain $a \in A$, $b \in B$, but not others.

So which is the correct interpretation?

Answer 1

If $f$ is a function from $X$ to $Y$ and $A \subseteq Y$ then $f^{-1}(A)$ is defined as $\{x \in X: f(x) \in A\}$. (Note that this does not require existence of the inverse function $f^{-1}$). From this definition it is obvious that $x \in f^{-1}(A)$ iff $f(x) \in A$.

In the definition of $A_0$ and $B_0$ we treat $a$ and $b$ as fixed. So 'for all $a$ and $b$' is not correct).

Answer 2

The points $a\in A$ and $b\in B$ are fixed points chosen at the beginning. The function $p$ is then defined in terms of them, and $A_0$ is simply $p^{-1}[A]$, the inverse image of $A$ under $p$. By definition this means that a real number $t$ belongs to $A_0$ if and only if $p(t)\in A$.

Yes, the function $p$ depends on the choice of $a$ and $b$: choose a different point $a_0\in A$ and a different point $b_0\in B$, and you get a different function $p_0:\Bbb R^1\to\Bbb R^k$. This does not matter, however, because the point of the theorem is that all of these functions have the property that the inverse images under them of $A$ and $B$ are separated sets in $\Bbb R^1$.

Answer 3

In the previous sentence, $ a $ and $ b $ are introduced, and everything that follows depends on them. So given $ A $, $ B $, $ a $, and $ b $, there is only one function $ p $ and only one set $ A _ 0 $, which is the preimage of $ A $ under that function $ p $. And the definition of preimage says that, given an element $ t $ of the domain of $ p $, $ t $ belongs to the preimage if and only if $ p ( t ) $ belongs to $ A $.

The logical structure of the paragraph is as follows:

We have $ \mathbb R $ generally.

Introduce a natural number $ k $.

Now we can talk about $ \mathbb R ^ k $.

Introduce $ A $ and $ B $ as separated subsets of $ \mathbb R ^ k $.

Introduce $ a $ as an element of $ A $ and $ b $ as an element of $ B $.

Introduce $ t $ as an element of $ \mathbb R ^ 1 $.

Now we can consider $ ( 1 - t ) a + t b $ in $ \mathbb R ^ k $.

Abstract over the above to define a function $ p $ from $ \mathbb R ^ 1 $ to $ \mathbb R ^ k $. (We no longer have $ t $.)

Define subsets $ A _ 0 $ and $ B _ 0 $ of $ \mathbb R ^ 1 $ as the respective preimages of $ A $ and $ B $ under $ p $.

Introduce $ t $ for purposes of the remark.

Remark that $ t \in A _ 0 $ if and only if $ p ( t ) \in A $. (This is basically the definition of preimage.)

End remark; we no longer have $ t $.

Now you need to prove that $ A _ 0 $ and $ B _ 0 $ are separated.

At this point, we still have $ k $, $ A $, $ B $, $ a $, and $ b $, as well as the things that were defined using them ($ p $, $ A _ 0 $, and $ B _ 0 $). So all of these can be referred to in your proof. You might want to introduce other variables in your proof (such as $ t $ once again), but you will have to state what they mean when you do.

When all is said and done, all quantifications are made, all contexts closed, your theorem has the form $$ \forall \, k \in \mathbb N , \; \forall \, A , B \subseteq \mathbb R ^ k , \; \operatorname { separated } ( A , B ) \; \Rightarrow \; \forall \, a \in A , \; \forall \, b \in B , \; \ldots $$ (where the dots are the part that you have to prove).