I am trying to prove that a set, $A$ of $\mathbb R^2$ with $x + y <10$, is convex. These are the steps I have gone so far:
Let says $(a, b) \in A$ and $(c, d) \in A$. Then I need to prove that for any numbers $m$ and $n$, $m+n = 1$:
$$\begin{align} n(a,b) + (1-n)(c,d) &<10\\ \big((na + (1-n)c), (nb+(1-n)d) \big) &<10. \end{align}$$
I first observe that $a+b < 10$ and $c+d<10$. Since $0 \leq n \leq 1$ and $0 \leq (1-n) \leq 1$, therefore
$$\begin{align} n(a+b) &\leq (a+b)\\ &\leq 10,\\ \end{align}$$ and also $$\begin{align} (1-n)(c+d) &\leq (c+d)\\ &\leq 10.\\ \end{align}$$
But then I am lost on what to do next. Any complete help would be very much appreciated, I am pretty much a beginner in this craft. I hope this posting is not a duplication since I didn't see any similar problem from the prompts. Thank you.
I'd recommend not getting bogged down in the details of each coordinate of a vector in $\Bbb R^2$. Instead, just let $x, y \in A$ (so they're vectors). This has the advantage of being able to easily extend to $\Bbb R^n$, rather than just $\Bbb R^2$.
We can rewrite $A$ as $A=\{x : x^\top 1 < 10\}$, where $1$ represents the vector of all ones. Now we aim to show, for any $t\in [0, 1]$, we have:
$$ x t + y(1-t) \in A $$
Then: \begin{align} (x t + y(1-t))^\top 1 &= (x t)^\top 1 + (y(1-t))^\top 1\\ &=t(x^\top 1) + (1-t)(y^\top 1) \\ &< t(10) + (1-t)(10) \\ &< 10 \end{align}
Thus, we have $xt + y(1-t)\in A$ for any $x, y \in A$ and $t\in [0, 1]$. Therefore, the set $A$ is convex, as was to be shown.