# How many 'supremum(s)' and 'infimum(s)' can a set have?

I am learning calculus/real analysis with Apostol's Calculus (2nd Edition). I have a doubt about the grammer of this book. Apostol, everywhere, uses a supremum (or a least upper bound) and an infimum (or a greatest lower bound) for some set $S$.

How many suprema and infima can a set have that he has used determinents like a and an instead of the.

As far as my learning is concerned, I was sure about the fact that supremum and infimum are unique for a given set.

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Supremum and infimum of a set are unique extended real numbers. – user21436 Jan 17 '12 at 11:00
@Kannappan Exactly this is what I know. PS: Kannappan edited their comment and my comment looks regardless. – gaurav Jan 17 '12 at 11:03
Correct usage of the articles a(n) and the can be tricky. The following are correct: (1) $A$ has a supremum. (2) The supremum of $A$ is $1$. ‘$A$ has the supremum’ by itself is incorrect, and ‘A supremum of $A$ is $1$’ is somewhat misleading. – Brian M. Scott Jan 17 '12 at 11:17
The set $\{a,a,a,a\}$ has exactly one element. For every anti-symmetric relation, suprema and infima are unique. – Michael Greinecker Jan 17 '12 at 11:20
‘$a$ is the supremum of $S$’ is fine. (I used $a$ instead of $A$ because the upper-case $A$ suggests a set rather than a single real number.) – Brian M. Scott Jan 17 '12 at 11:28

Let $(X,\leq)$ be a partially ordered set. That means that for all $x,y,z\in X$:

1) $x\leq x$ (reflexivity)

2) $x\leq y$ and $y\leq x$ imply $x=y$ (anti-symmetry)

3) $x\leq y$ and $y\leq z$ imply $x\leq z$ (transitivity)

An element $x\in X$ is a lower bound of a subset $S\subseteq X$ if $x\leq S$ for all $s\in S$. An element $x$ is the greatest element in a subset $T\subseteq X$ if $x\in T$ and $t\leq x$ for all $t\in T$. An infimum of a set $S$ is a greatest element in the set of lower bounds of $S$. We will show that there can be at most one greatest element in every set, so there can be at most one infimum for every set.

There can be at most one greatest element in a subset $T\subseteq X$.

Proof: Let $x,x'$ be both greatest elements in $T$. Then $x\in T$ and $x'\in T$ and since $x$ is a greatest element in $T$, we have $x'\leq x$. Similarly, $x\leq x'$. By anti-symmetry, $x=x'$.

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So I should stick to the fact that sup. and inf. are unique for any real set. – gaurav Jan 17 '12 at 11:45
Yes, you should. – Michael Greinecker Jan 17 '12 at 12:01

In general, you can have more than one infimum. Consider sets of pairs of numbers $(a,b)$ and the relation on pairs $(a,b) \succeq (c,d)$ iff $a\ge c$ and $b\ge d$. This is only a partial order, and so, with respect to this relation, the following set has two distinct infima: $\{(1,2),(2,1),(2,2)\}$. That is, there are two elements that satisfy: "no element of the set is less that this element".

If the set is totally ordered (like the reals) then the infimum and supremum are unique.

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What you call suprema and infima are usually known as maximal and minimal elements. The common definition of an infimum is "the largest lower bound" and a supremum is the "smallest upper bound". In particular, an infimum is comparable with every other lower bound (similar for suprema), which means, by anti-symmetry of every partial order, that an infimum, if it exists, is unique. – Michael Greinecker Jan 17 '12 at 11:29
This is incorrect. If $\langle P,\preceq\rangle$ is a partial order, the infimum $m$ of a set $S\subseteq P$ must satisfy two conditions: $m\preceq s$ for all $s\in S$, and if $t\preceq s$ for all $s\in S$, then $t\preceq m$. In other words, it must be the greatest lower bound. Your partial order has no infimum, though it has two minimal (not minimum) elements. – Brian M. Scott Jan 17 '12 at 11:32
Sometimes it is very odd to say 'a largest' and 'a smallest'. 'the' is written before superlative degree of adjectives. – gaurav Jan 17 '12 at 11:36
I think I was taught that the first condition for $s$ being a supremum was $\neg (m \succ s)$. But perhaps I'm wrong. Anyway, the answer is sensitive to the definition of supremum. – Seamus Jan 17 '12 at 12:16