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I'm trying to complete this problem:

Let $A$ be a nonempty set and suppose $\alpha$ and $\beta$ are both suprema of $A$. Prove that $\alpha = \beta$.

The first thing i did was try to find an example, because i just simply am unsure where to start such a proof, since this is all quite intimidating when first learning analysis.

I came across a proof that proves the supremum of a set $A$ is unique if it exists: https://www.math.ucdavis.edu/~hunter/m125b/ch2.pdf . Where the author says:

Suppose that $M, M′$ are suprema of $A$. Then $M ≤ M′$ since $M′$ is an upper bound of $A$ and $M$ is a least upper bound; similarly, $M′ ≤ M$, so $M = M′$.

Ive tried quite a bit of searching, and it may be the case where i couldnt find an answer, because im not even sure what to call my question , if that makes sense? It feels like a "dumb" question, but i just simply dont understand the logic of the statement above, or basically that:

If $x,y \in \mathbb{R}$ where $x\leq y$ and $y\leq x$. Does $x=y$?

Sorry if this is a repost.

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  • $\begingroup$ Yes, the justification probably follows easily from the definition of $\leq$, whatever it may be. $\endgroup$
    – Git Gud
    Apr 14, 2015 at 22:07

2 Answers 2

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$\le$ is a total order on $\mathbb{R}$. Once $x\le y$, then either $x<y$ or $x=y$. Once $y\le x$, then either $y<x$ or $x=y$. By the trichotomy principle, exactly one of $x<y$, $x=y$, or $x>y$ must hold. Combining these three ideas we conclude $x=y$.

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If a set has a minimum, then this minimum is unique.

Indeed if $S$ has an element $m$ such that $m\le s$, for all $s\in S$ and $m'$ is another element with the same property, then $m\le m'$ and $m'\le m$. So $m=m'$ (the order relation is antisymmetric).

The supremum of a set $A$ is the minimum of the set of upper bounds of $A$ (provided it exists).

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