TITLE: Prove that the interval $[a,b]$ in $\Bbb R$ is the same as the segment $[a,b]$ in $\Bbb R^1$
QUESTION [5 upvotes]: Prove that the interval $[a,b]$ in $\Bbb R$ is the same as the segment $[a,b]$ in $\Bbb R^1$. 
That is,
$\{x\in\Bbb R: a\le x\le b\} = \{y\in\Bbb R: \exists s,t\in [0,1], s+t=1\text{ and } y=sa+tb\}$
I was trying to solve it as follows. Let's denote 
$A= \{x\in\Bbb R: a\le x\le b\}$ and 
$B= \{y\in\Bbb R: \exists s,t\in [0,1], s+t=1\text{ and }y=sa+tb\}$. Then I wanted to show, that $A\subseteq B$ and $B\subseteq A$ so $A = B$.
For the $B\subseteq A$ I did the following:
$y=sa+tb=sa+(1-s)b=s(a-b)+b$ that is obviously in $[a,b]$ (as when $s=0 $ then $y=b $ and when $s=1 $ then $ y=a$). So $B\subseteq A$.
For $A\subseteq B$ I tried to show that if $x\in [a,b] $ then we can express $x=a+(b-a)k, k\in [0,1]$ 
then $x=a(1-k)+bk$
Saying that $(1-k)=t $ and $ k=s$ we can conclude that $A\subseteq B$. 
It seems to me that the second part is incorrect, but, unfortunately I can't figure out what should be done here.

REPLY [2 votes]: Note: the following is a proof. If you desire to prove it on your own, I recommend looking through the comments above without reading this post.
To show that $B\subseteq A$, simply notice that for all $s,t\in[0,1]$ with $s+t=1$ we have 
$$ a = (s+t)a = sa+ta \le sa+tb \le sb+tb = (s+t)b = b,$$
thus $sa+tb\in A$. Hence $B\subseteq A$.
To show that $A\subseteq B$. Take any $x$ with $a\le x\le b$. Then let $\lambda = \frac{x-a}{b-a}\in[0,1]$. Then notice that $1-\lambda = \frac{b-x}{b-a}$ and so
$$ (1-\lambda) a + \lambda b = \frac{b-x}{b-a}\cdot a + \frac{x-a}{b-a}\cdot b = \frac{ba-xa+sb-ab}{b-a} =\frac{(b-a)x}{b-a}=x. $$ 
Letting $s=1-\lambda$ and $t=\lambda$, it is clear that $x=sa+tb\in B$. Hence $A\subseteq B$. Putting these things together we find that $A=B$.