teacher said that this set is bounded without any elaboration,to me it is not obvious that it is bounded,in fact i think that it is unbounded and now i am confused. $$S=[{(x,y,z)\in R^3}|x\geq 0,y\geq 0,z\geq 0,x+y+z=2]$$ Now this is a plane in the first octant and i think that it is unbounded since there isn't any M that can bound $$x+z+y-2$$.I don't understand why it is bounded.Thanks in advance.
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of course there is no bound for $x+y+z-2$, but for $x+y+z-2=0$ there is. – Arun May 25 '17 at 08:55
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Its clear that if any of $x,y,z$ exceed 2, the equality cannot be satisfied (due to the non-negativity constraint). Thus, you can see its a subset of $[0,2] \times [0,2] \times [0,2]$ which is clearly bounded.
Batman
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One more thing:If we were to let negative values in,would it still be bounded?If for example we had x>=-1 instead of x>=0,would it still be bounded? – DrStrange May 25 '17 at 09:05
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Yes. Say $x,y,z \geq -1$. Then, you'd have the bounds $[-1,3] \times [-1,3] \times [-1,3]$. – Batman May 25 '17 at 09:13