Prove $\{x \in \mathbb{R}:a\leq x\leq b\}=\{y\in \mathbb{R}:\exists s,t\in [0,1]\; with\; s+t=1\; and\; y=sa+tb\}$

Question

I'm doing the $\supseteq$ direction first as I find that the easiest.

Let $A={\{x \in \mathbb{R}:a\leq x\leq b}\}$

Let $B=\{y\in \mathbb{R}:\exists s,t\in [0,1]\; with\; s+t=1\; and\; y=sa+tb\}$

Suppose y is an arbitrary element of B. Then $\exists s,t\in [0,1]\; with\; s+t=1\; and\; y=sa+tb$

then $sa+ta=a$ and $sb+tb=b$.

Since $a\leq b$, $tb\geq ta$ then $sa+tb\geq a$

And $sa\leq sb$ so $a\leq sa+tb \leq b$

Since $y=sa+tb$ then $y \in [a,b]$ as required.

For the $\subseteq$ direction I feel I didn't do this right.

Let A,B as above. Suppose x is an arbitrary element of A. Then $a\leq x\leq b$.

Suppose $\exists s,t\in [0,1]\; with\; s+t=1$.

then $a=sa+ta \; and \; sb+tb=b$

Since $a\leq b$, $sa\leq sb \; and \; ta\leq tb$

then $a=sa+ta\leq sa+tb\leq sb+tb=b$

then $sa+tb$ is an arbitrary element of A. Thus $x=sa+tb$ and $x\in B$.

Answer 1

The line "Suppose $\exists s,t\in[0,1]$ with $s+t=1$" appears to be your first error.

You don't need to suppose there exist such an $s$ and $t$, they clearly exist. You can let $s=\frac{1}{2}=t$, then $s,t\in[0,1]$ and $s+t=1$. Therefore there exist $s$ and $t$ that satisfy the conditions that you request.

Unfortunately, $s$ and $t$ won't satisfy the other condition that you haven't stated, namely that $as+bt=x$.

What you need to do is to construct the appropriate $s$ and $t$.

Hint: Do some scratch work (in other words, figure out what $s$ and $t$ have to be using some side work, then just pick the right numbers out of thin air). Scratch work is not part of the final answer, it's just some work to figure out what the right form of the answer should be. Since it's not part of the final answer, we can assume anything we want in scratch work.

Scratch Work:

So, if you assume that $s+t=1$ and $x=as+bt$, then $x=as+b(1-s)$. From here, you get $s=\frac{b-x}{b-a}$. Similarly, you can get an expression for $t$.

Now, in the proof, you pick the right values of $s$ and $t$ "as if by magic", show that they have the right properties (they are between $0$ and $1$, and sum to $1$) and $as+bt$ equals $x$.

Answer 2

One direction is straightforward.

We have $a = sa+ta \le sa+tb \le sb+tb = b$.
Hence $sa+tb $ is in the first set (A).

The other direction just needs a little trick, suppose $x$ is in the first set, let $s = {x-a \over b-a}$, $t = {b-x \over b-a}$. Then $s,t$ are non negative, $s+t = 1$ and $x = sa+tb$. Hence $x$ is in the second set.