Show that interior of C is convex set

Question

A set $C$ in euclidean space is said to be convex if for any $x$ and $y$ in $C$, $tx + (1-t)y$ belongs to $C$ for any $t$ between $0$ and $1$.

Show that the interior of $C$ is convex.

Appreciate your help!

Answer 1

It's because you can construct a neighbourhood of $tx+(1-t)y$ by using neighbourhoods of $x$ and $y$ and interpolating. If you're talking about $\mathbb R$ you replace "neighborhood" with an open interval, if you're talking about $\mathbb R^n$ or similar you can replace "neighborhood" with a ball, (neighborhood is the generic concept, a open set containing the point in question):

Since $\operatorname{int}C\subseteq C$ it's clear that $a=tx+(1-t)y \in C$.

Now since $x$ and $y$ are in the interior of $C$ there is a congruent neighbourhood round them of equal (we can create a neigbourhood of each and translate to the origin and join them). That is that for each $\delta \in B$ we have that $x+\delta$ and $y+\delta$ is in $C$ and therefore also $t(x+\delta)+(1-t)(y+\delta) = tx+(1-t)y+\delta = a+\delta$. But $\{a+\delta: |\delta|<r\}$ is a ball round $a$ within $C$.

To clarify how $B$ is constructed. Let $\Omega_x$ and $\Omega_y$ be neighborhoods of $x$ and $y$ then the set $V_x = \Omega_x-x = \{v : v+x \in \Omega_x\}$ and $V_y = \Omega_y-y$ are neighborhoods of origin and $B=V_x \cap V_y$ is too. $B+x$ is therefore a neighborhood of $x$ and $B+y$ is of $y$.