Let $A ⊂ \Bbb R^2$ be a fixed convex set and let $X_1, \cdots , X_n ⊂ \Bbb R^2$ be any convex sets such that every three of them intersect a translation of $A$. Then there exists a translation of $A$ that intersects all sets $X_i$.
The proof of the same is given as follows:
So what I have understood is this as given below:
But how did they assume that $Y_i$ is a convex set? Don't we have to prove that it is a convex set? I want to know how to prove $Y_i$ as convex set.
i) If $a$ is any interior point in the triangle $A$, then $A'$ is a translations and $a'$ is a corresponding point (We also use the notation $A''$ for translations)
Fix a line segment $[pq]$ Then define $D_{[pq]}^A = \bigcup_{a'\in [pq]}\ A'$
That is, we can view $D_{pq}^A$ as a trajectory when we push $A$ along $[pq]$. Hence the set is convex.
If $a'=p,\ a''=q$ and $A'\bigcap A''$ is nonempty, then
ii) Fix a point $r$ so that we define $$ R = \bigg\{ a' \bigg| A'\bigcap \{r\} \neq \emptyset \bigg\} $$
If $a',\ a''\in R$, then $[a'a'']$ is in $R$ by step i) so that $R$ is a convex set.
iii) When $X$ is a convex set, we define $Y = \bigcup_{p\in X}\ P$.
Here note that $Y$ coincides to the set defined in OP.