Supremum Proof Analysis

Question

Suppose that $x_n \leq y_n, \forall n$. Show that $\sup x_n \leq \sup y_n$. I would like to prove this by contradiction but am slightly stuck as how to go about doing this. I started off with sssuming that $$\sup x_n = x > \sup y_n = y.$$ Then we have that $\forall 0 < \epsilon < x - y$, $\exists N$ such that $\forall n \geq N$, we have $$0 < y - y_n < \epsilon < x - y.$$ I am stuck here and any help would be greatly appreciated.

Answer 1

Instead of $y-y_n$, consider $x-x_n$.

Then we have for some $N$: $$0\le x-x_N<\epsilon<x-y$$

which gives $y<x_N$.

By definition of $\sup$, $y_n\le y \,\forall n$, and thus $y_N\le y<x_N$.

This is a contradiction.

Answer 2

I don't think you need contradiction.

We have $x_n \leq y_n$ for all $n$. Let $y^* = \sup_n y_n$. Then $y^* \geq y_n \geq x_n$ for all $n$. Let $x^* = \sup_n x_n$. Then the supremum is the least upper bound, so $x^* \geq x_n$ for all $n$ and if there is any other value $z$ so that $z \geq x_n$ for all $n$, then $z \geq x^*$. Notice $y^*$ fits the bill of $z$ here, so $x^* \leq y^*$.

Contradiction also works though.

Answer 3

Here's a completely alternative approach, but it can apply in more cases.

Suppose $X, Y \subset \mathbb{R}$ have the property that, for each $y \in Y$ there exists $x \in X$ with $x \geq y$. Suppose $\sup X \ne \sup Y$, and for contradiction assume that $\sup X < \sup Y$. Then we can fix $y \in Y$ with $\sup X < y \leq \sup Y$, and furthermore $x \in X$ satisfying $y \leq x$. But this means $\sup X < x$, a contradiction. Therefore, $\sup Y = \sup X$ or $\sup Y < \sup X$, which is a tedious way to say that $\sup Y \leq \sup X$.

Our case involves taking $X = \{x_n : n \in \mathbb{N}\}$ and $Y = \{y_n : n \in \mathbb{N}\}$. Note that our property holds because for each $y_n \in Y$, $x_n \geq y_n$.

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In fact, for any $X, Y \subset \mathbb{R}$, $\sup Y \leq \sup X$ if and only if the above holds.

Answer 4

You have to get and understand what the definitions mean and develop an "yeah, that's obvious" sense (and such an intuition can only come with practice and rereading and rereading the definitions and drawing lot of pictures).

single line proof: If $\sup x_n > \sup y_n$ then $\sup y_n$ can not be an upper bound of $\{x_n\}$ so there is an $x_k > \sup y_n$. So $x_k > \sup y_n \ge y_k$. A contradiction.