Proving the greatest lower bound. 
Define sets of real numbers $X$ and $Y$ by $X= \{x_1, x_2,\ldots ,x_n\}$ and $Y=\{y_1, y_2, \ldots,y_n\}$ for some $n\in \Bbb{N}$.
Define $X+Y$ by $X+Y=\{x_i+y_i: 1\le i \le n\}$.

*

*Explain why $X+Y$ must be bounded below.


I'm not sure how to begin to answer this



*Prove $\operatorname{inf}(X+Y) \ge \operatorname{inf} X + \operatorname{inf} Y$.


Here is my proof:
First let $Z$ denote the set $Z= X+Y=\{x_i+y_i: 1\le i \le n\}$. If $X$ and $Y$ have infimum, then $Z$ has an infimum and $\operatorname{inf} X= \operatorname{inf} X+  \operatorname{inf} Y$. Let $\operatorname{inf} X=x$ and $\operatorname{inf} Y=y$ and also let $z\in Z$. Then $a=x+y$ for some $a\in X$ and $b\in Y$. Thus $z=a+b\ge x+y$ so $x+y$ is a lower bound of $Z$. By the completeness axiom, $Z$ has a greatest lower bound such that $Z=c$. Since $c$ is the greatest lower bound of $Z$, then $c\ge a + b$.
Is this a decent proof? Please advise.
 A: Since it was tagged as proof-verification, to answer this question, I feel the need to proofread the proposed proof in the question.


*

*Since $X$ and $Y$ (i.e. $X+Y$) are finite, you can choose the minimum of the set $X+Y$ to be a lower bound for $X+Y$.  For the general (non-finite) case, use the result of the next question to give a positive answer.

*
If $X$ and $Y$ have infimum, then $Z$ has an infimum.

In the question, the sets $X,Y$ are finite, so it's correct, but if we drop this condition, which is unnecessarily strong, we can't make such "if-then" deduction in the proof.
The equality

$\inf X= \inf X+  \inf Y$

is, in general, incorrect, not even if the LHS is changed to $\inf Z$.  (Exercise: find a counterexample using finite sets to illustrate this.)  You may remove this assertion, while keeping $x=\inf X,y=\inf Y$.  Then you let $z\in Z$, so it should be $z=a+b$ for some $a\in X$ and $b \in Y$, (IMHO, it's better to denote the arbitrary element in $Z$ with another alphabet that differs "to a greater degree" than the fixed infimums $x,y$.  It looks better, but since this doesn't affect the logic, I'll keep using $z$.) so that the inequality $z=a+b\ge x+y$ holds.  The rest of the proof is fine, **except the last inequality

$c\ge a+b$

It should be $c \ge x+y$ instead since you've said that $x+y$ is a lower bound for $X+Y$ and $c$ is the greatest among those lower bounds.  This completes your proof.

A second writing

$\inf (X+Y) \ge \inf X + \inf Y$

Justification: Let $x\in X,y \in Y$.  By the very definition of infimum, we have
\begin{align}
\inf X \le& x \\
\inf Y \le& y
\end{align}
Add them together to get $$\inf X+\inf Y\le x+y.$$
Since the choice of $x,y$ are arbitrary, $\inf X+\inf Y$ is a lower bound for the set $X+Y$, and it is smaller than the greatest lower bound for $X+Y$.  That is, $$\bbox[2px, border:1px solid black]{\inf X+\inf Y \le \inf(X+Y)}$$
N.B. This proof works for sets $X$ and $Y$ indexed by arbitrary index sets $I$ and $J$ respectively.  ($I$ can be different from $J$.)  This reveals the true potential of the inequality in part (2).
