Assume that $K\subseteq \mathbb{R}^n$ is convex and $x_1,...,x_k\in K$. Show that any convex combination of $x_1,...,x_k$ is contained in $K$.
Relevant definitions:
Definition of Convexity: A set $k\subseteq\mathbb{R}$ is convex if for all $x,y\in k$, the line segment between $x$ and $y$ is entirely contained in $k$.
Definition of convex combination: A convex combination of $x_1,...,x_k\in\mathbb{R}^n$ is $$x=\lambda_1x_1+\lambda_2x_2+...+\lambda_kx_k$$ such that $0\leq\lambda_i\leq1$ and $\lambda_1+\lambda_2+...+\lambda_k=1$
My Attempt:
We can solve this by induction on $k$.
Base Case: By the definition of convexity, we know that the statement is true for $k=2$.
Induction Hypothesis: Assume that the statement holds for some value $l$. i.e. For $x_1,...,x_l\in K$, and $\lambda_1,...,\lambda_l\geq0$ with $\lambda_1+...+\lambda_l=1$, we have $\lambda_1x_1+...+\lambda_lx_l\in K$.
Induction Step: Let $x_1,...x_l,x_{l+1}\in K$ and $\lambda_1,...,\lambda_l,\lambda_{l+1}\geq0$ with $\lambda_1+...+\lambda_l+\lambda_{l+1}=1$. Now, assume that $\lambda_1\neq 1$. Then $$y=\lambda_1x_1+(1-\lambda_1)(\mu_2x_2+...+\mu_lx_l+\mu_{l+1}x_{l+1})$$ where $\mu_i=\frac{\lambda_i}{1-\lambda_1}\geq0$. Then $$\mu_2+...+\mu_{l+1}=\frac{\lambda_2+...+\lambda_{l+1}}{1-\lambda_1}=1$$ Since $K$ is convex and $x_2,...x_{l+1}\in K$, we have that $\mu_2+,...\mu_{l+1}x_{l+1}\in K$. Then since $x_1\in K$, we have that $y\in K$
I don't think I did this correctly because I couldn't find where I needed to apply the induction hypothesis. Any pointers would be very helpful. Thanks.
