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I have a question.

I have to show that $$S1 = \{x \in \mathbb{R}^2 : x_1 \geq 0,x_2 \geq 0, x_1 + x_2 = 2\}$$

is a bounded set.

So I have to show that $\sqrt{x1^2+x2^2}0$ and we have $0 \leq x_1 \leq2$ and $0 \leq x_2\leq2$. And $x_2 = 2-x_1$

We can fill in $\sqrt{x_1^2 + (2-x_1)^2} = \sqrt{0^2 + (2-0)^2} = 2 < M = 3$. And we can fill in $\sqrt{x_1^2 + (2-x_1)^2} = \sqrt{2^2 + (2-2)^2}= 2 < M = 3$.

Every value between the 0 and the 2 that satisfy $x_1+x_2 = 2$ is smaller than this $M$. So the set is bounded.

Is this correct?

  • 0
    You know, on this site, it's traditional for you to ask questions, then after you get an answer, to provide some feedback. You can do that by commenting, upvoting the questions and, most importantly, *accepting* questions that are useful to you. That allows us to keep the page tidy as your questions are then marked as finished. – 2017-01-30
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    Did you try to draw the set $S1$? – 2017-01-30

2 Answers 2

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To show it is bounded it is enough to draw a circle in the plane around it. The set is a segment of a line as it passes through quadrant 1, so it should be easy to get the endpoints, then the midpoint (to be used as the center of a circle), then an appropriate radius (maybe half the distance between the ends, or could use that plus $1$ to make sure the set is interior to the circle, if that's needed.

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For $(x_1,x_2) \in S_1$ we have

$||(x_1,x_2)||^2 =x_1^2+x_2^2 \le x_1^2+2x_1x_2+x_2^2=(x_1+x_2)^2=4$

Hence $||(x_1,x_2)|| \le 2$