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Let $C$ be a closed convex set in $\Bbb R^2$.

Define $C_x=\{y:x+ty\in C\forall t>0\}$.

Show that if $x,x^{'}\in C$ then $C_x=C_{x^{'}}$.

Attempt:

Since $x,x^{'}\in C$ we have a continuous path $f:[0,1]\to C$ such that $f(0)=x,f(1)=x^{'}$

Let $y_1\in C_x\implies x+ty_1\in C$.

To show that $x^{'}+ty_1\in C_{x^{'}}$.

I need a path in $C$ which joins $x^{'}+ty_1$ with some point in $C$.

Will you please give some hints on how to find such a path?

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    Draw a picture. Choose distinct $x, x^\prime$ in $\mathbb{R}^2$, and choose some fixed direction for $y$. The points $x + ty$, $t > 0$, form a ray emanating from $x$. Draw that ray and the ray parallel to it from $x^\prime$. Now draw all segments from $x^\prime$ to the points on the ray emanating from $x$, and consider the points of those segments which lie on the segment from $x + t_1y$ to $x' + t_1y$. The point you _want_ to get is $x' + t_1y$. Although you don't get it directly, you do get close. But $C$ is closed, so ... – 2017-01-17
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    But I can't write this answer "Draw a picture.." in exam;How can I present it – 2017-01-17
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    All I gave was hints. The intersection points can be found algebraically, and then the resulting limit clinches it. – 2017-01-17
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    The picture guides you; the corresponding algebra makes it rigorous. – 2017-01-17

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