Let $X,Y$ be two bounded subsets of $\mathbb{R}$ satisfying the following proposition 1 :
$$\forall x \in X, \forall y \in Y : x \leq y $$

I wanted to know if there's a direct proof of $$\sup X \leq \inf Y$$

I think I managed to prove it by contradiction ( correct if I'm wrong please ).

Suppose $\sup X > \inf Y$ . It then follows that $\sup X - \inf Y > 0$ .

Choosing $\varepsilon_1 = \sup X - \inf Y,$ by definition of infimum, we have that $$\exists y_0 \in Y [ y_0 < \inf Y + ( \sup X - \inf Y ) = \sup X ] $$

Choosing $\varepsilon_2 = \sup X - y_0, $ by definition of supremum, we have that $$\exists x_0 \in X [ (\sup X - (\sup X - y_0) = y_0 < x_0 ] $$
Last statement contradicts proposition 1.

I always try to avoid proving by contradictions, but regarding this proposition I can't a direct proof.

Thanks a lot.


You can argue without $\epsilon$s:

If $B$ is an upper bound for $X$, then you must have $\sup X \le B$.

Since you have $x \le y$ for all $x \in X, y \in Y$, each $y$ is an upper bound for $X$, that is, we have $\sup X \le y$ for all $y \in Y$.

Similarly, if $C$ is a lower bound for $Y$, we must have $C \le \inf Y$, and since $\sup X$ is a lower bound for $Y$, we have $\sup X \le \inf Y$.

  • $\begingroup$ Really quick and clean. Thanks. $\endgroup$ – nerdy Aug 25 '14 at 21:25

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