Hello! I've got some questions concerning convex sets.(adsbygoogle = window.adsbygoogle || []).push({});

We've had a lecture about convex sets this week, and got a some basic problems to solve. I think I can't use the material in the lecture to solve the problem. I'm just not sure about whether I fully understand the concept and can use it properly.

In the lecture the prof presented a proof for the fact that for every [itex]t\in\matbb{R}[/itex] the interval [itex]I=[0,t]:=\{s|s\in\mathbb{R},0\le{s}\le{t}\}[/itex] is convex.

The proof went like this.

Let [itex]s_1,s_2\in[0,t][/itex] and [itex]\lambda\in[0,1][/itex]. The following holds: [itex]0\le{s_1}[/itex] and [itex]s_2\le{t}[/itex].

[itex]\lambda{s_1}+(1-\lambda){s_2}\le\lambda{t}+(1-\lambda)t=t[/itex]. Hence, the interval is convex.

Now, the first question.

Why cannot we simply check whether the condition holds for [itex]\lambda=0[/itex] or [itex]\lambda=1[/itex] ?

It's [itex]\lambda\cdot{0}+(1-\lambda)t[/itex]. Then for [itex]\lambda=0[/itex] t is in I and for [itex]\lambda=1[/itex] 0 is in I.

It's now asked to show that a hyperplane [itex]E\subseteq\mathbb{R}^n[/itex], [itex]E:=\{(x_1,...,x_n)\in\mathbb{R}^n|a_1{x_1}+...+a_n{x_n}=b\}[/itex], where [itex]a_1,...,a_n,b\in\mathbb{R}[/itex] and [itex](a_1,...,a_n)\neq(0,...,0)[/itex] is a convex set.

Here, equally, it seems too easy just to put in two elements, say [itex]x_1[/itex] and [itex]x_n[/itex] and let [itex]\lambda[/itex] be either 0 or 1. If it's wrong to take this way, how can I show that E is convex differently?

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# Convex sets

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