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.

  • 2
    $\begingroup$ The sets are finite, so the $\inf$'s are actually $\min$'s. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 9 '17 at 23:52
  • $\begingroup$ Please do not delete questions after having received an answer. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 28 '19 at 11:00
  • $\begingroup$ 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). $\endgroup$ – 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).


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