Show that the interior of a convex set is convex Question: Let $P\subseteq \mathbb{R}^n$ be a convex set. Show that $\text{int}(P)$ is a convex set.
I know that a point $x$ is said to be an interior point of the set $P$ if there is an open ball centered at $x$ that is contained entirely in $P$. The set of all interior points of $P$ is denoted by $\text{int}(P)$.
Also, to say that a set $P$ is convex means that if $x,y \in P$ then $tx+(1-t)y \in P$ for all $t \in (0,1)$.
How to go about the above proof?

I'll give a proof based on the following picture:

Suppose that $x$ and $y$ are interior points of a convex set $\Omega \subset \mathbb R^n$. Let $0 < \theta < 1$. We wish to show that the point $z = \theta x + (1 - \theta) y$ is in the interior of $\Omega$.
There exists an open ball $A$ centered at $x$ such that $A \subset \Omega$. Let $r$ be the radius of $A$, and let $B$ be the open ball of radius $\theta r$ centered at $z$.
Claim: $B \subset \Omega$.
Proof: Let $\hat z \in B$. There exists a point $\hat x \in \mathbb R^n$ such that
$$
\hat z = \theta \hat x + (1 - \theta) y.
$$
I will show that $\hat x \in A$. It will follow, by the convexity of $\Omega$, that $\hat z \in \Omega$. Since $\hat z$ is an arbitrary point in $B$, this will show that $B \subset \Omega$.
To complete the proof, note that
\begin{align}
\| \theta(\hat x - x)\| &= \| \theta \hat x - \theta x \| \\
&= \| \hat z - (1 - \theta) y - (z - (1 - \theta) y) \| \\
&= \| \hat z - z \| \\
&\leq \theta r.
\end{align}
It follows that $\| \hat x - x \| \leq r$, which shows that $\hat x \in A$.
