Check if the set $A=\{(x_{1},x_{2},x_{3})\in \mathbb{R}^{3}: |x_{1}|+|x_{2}|+|x_{3}|\leq 1 \}$ is convex, closed, and bounded I need to check whether the set $$ A=\left\{(x_{1},x_{2},x_{3})\in \mathbb{R}^{3}: |x_{1}|+|x_{2}|+|x_{3}|\leq 1 \right\}$$ is convex, closed, and bounded.
To check if it is convex, I let $x^{(1)},x^{(2)} \in A$, $\lambda \in (0,1)$, and define $z \in \mathbb{R}^{3}$ as 
$$z = |(1-\lambda)x_{1}^{(1)}+\lambda x_{1}^{(2)}| +|(1-\lambda)x_{2}^{(1)}+\lambda x_{2}^{(2)}| +|(1-\lambda)x_{3}^{(1)}+\lambda x_{3}^{(2)}| $$
Then, I did the following series of algebraic manipulations to it:
$$z = |(1-\lambda)x_{1}^{(1)}+\lambda x_{1}^{(2)}| +|(1-\lambda)x_{2}^{(1)}+\lambda x_{2}^{(2)}| +|(1-\lambda)x_{3}^{(1)}+\lambda x_{3}^{(2)}| \\ \leq  |(1-\lambda)x_{1}^{(1)}|+|\lambda x_{1}^{(2)}| +|(1-\lambda)x_{2}^{(1)}|+|\lambda x_{2}^{(2)}| +|(1-\lambda)x_{3}^{(1)}|+|\lambda x_{3}^{(2)}|\\ =  (1-\lambda)\,|x_{1}^{(1)}|+\lambda\, |x_{1}^{(2)}| +(1-\lambda)\,|x_{2}^{(1)}|+\lambda\,  |x_{2}^{(2)}| +(1-\lambda)\,|x_{3}^{(1)}|+\lambda\, |x_{3}^{(2)}| \\ =  (1-\lambda)\,(|x_{1}^{(1)}|+ |x_{2}^{(1)}| +|x_{3}^{(1)}|)+\lambda\,  (|x_{1}^{(2)}|+|x_{2}^{(2)}|+ |x_{3}^{(2)}|) \\ \leq (1-\lambda)\cdot 1+ \lambda \cdot 1\\ = 1-\lambda + \lambda =1  $$
Therefore, the set is convex.
My first question is: Did I do this right? And if not, how do I fix what I did to make it right?
Now, to show that the set is closed, the $\leq 1$ seems a dead giveaway, but I actually have to prove it formally. I was thinking I could use the fact that the absolute value function is continuous, so then, if I had a convergent sequence of points in $\mathbb{R}^{3}$, then I could conclude that the absolute value of that sequence also converged to a point in the set $A$. However, I have never done this in $\mathbb{R}^{3}$ - I have only ever showed continuity in $\mathbb{R}^{2}$, so I don't know what this would "look like" - the algebra of it and all. I have two other problems like this to do, so I think that seeing this part done out completely in detail could help me figure out how to do the rest of the problems on my own.
As for the bounded part, the graph of $A$ is a type of polyhedron (I'm not sure exactly how many faces it has because it is hard to sketch), so since I can contain it inside a ball, centered at the origin, of radius, say, $2$, I suppose I could say it was bounded. Is this correct and/or formal enough?
I thank you very much for your time and patience!
 A: Basically this is the volume of the $ {L}_{1} $ Unit Ball in $ \mathbb{R}^{3} $.  
Since all norms are Convex (This is one property of a norm, derived from the Triangle Inequality) the above set is convex.
A: As Royi said, $A$ is the unit ball in $L_1$ norm. Since all norms in $\Bbb R^3$ are equivalent (they produce the same topology) and the ball in the Euclidean norm is compact, then $A$ is also compact (so closed and bounded). But be careful: if a normed space is infinite dimensional (not the case here), then the unit ball is not compact, nevertheless it is also closed and bounded.
In any norm the ball is a convex set which follows easily by a trangle inequality.
A: Your proof of convexity is okay, more or less. What you need to do is pick two arbitrary points $x^{(1)}$ and $x^{(2)}$ in $A$, then show every point $(1-\lambda)x^{(1)}+\lambda x^{(2)}$ along the line between $x^{(1)}$ and $x^{(2)}$ is in $A$. To do this you must show that the co-ordinates $(z_1,z_2,z_3)$ of  $(1-\lambda)x^{(1)}+\lambda x^{(2)}$ satisfy $|z_1|+|z_2|+|z_3| \le 1$. This is what your algebraic manipulation shows.
You're on the right track with using the continuity of the absolute value function. To be more explicit: take a convergent sequence $x^{(n)}$ in $A$ converging to some $x \in \mathbb R^3$. Every term satisfies $|x^{(n)}_1| + |x^{(n)}_2| + |x^{(n)}_3| \le 1$. Since the absolute value is continuous, this implies $x$ also satisfies this relation. So $x \in A$.
The bounded reasoning looks correct. To understand what the graph of $A$ looks like - think of the analogous case in $\mathbb R^2$. This is a diamond (a square rotated by 45 degrees). Then you can see it fits inside a ball of radius 1. The 3-dimensional case is the same. If you want to be more formal, you can prove that $A \subset B(0,1)$ by showing $|x_1|+|x_2|+|x_3| \le 1$ implies $\sqrt{x_1^2+x_2^2+x_3^2} \le 1$.
