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 $$X+Y=\{x_i+y_i: 1\le i \le n\}$$.

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

I'm not sure how to begin to answer this

1. 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.

• The sets are finite, so the $\inf$'s are actually $\min$'s. – GNUSupporter 8964民主女神 地下教會 Dec 9 '17 at 23:52
• Please do not delete questions after having received an answer. – GNUSupporter 8964民主女神 地下教會 Feb 28 '19 at 11:00
• This is a horrible notion of $X+Y$, as it is sensible to permutations of the $x_1, x_2, \ldots, x_n$ (and thus does not just depend on $X$ and $Y$ alone). – darij grinberg Mar 5 '19 at 18:03

Since it was tagged as , to answer this question, I feel the need to proofread the proposed proof in the question.

1. 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.
2. 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).