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I have a quick notational question.

Would correct logical notation for suprema and infima of some set $A \subset \mathbb{R}$ be:

$$\sup A = s : \forall a \in A : \forall b \ge a : a \le s \le b$$

$$\inf A = s : \forall a \in A : \forall b \le a : b \le s \le a$$

I could not find results quite answering this on MSE.

And, I am taking an advanced calculus or introduction to analysis course that uses Abbott's Understanding Analysis.

J. W. Tanner
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isaac
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1 Answers1

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Let $A = [0, 1]$. We have that $\sup A = 1$. However, it is not true that $\forall a \in A : \forall b \geq a : a \leq s \leq b$.

Take, for example, $a = 0, b = \frac{1}{2}$. Both $a \in A$ and $b \geq a$ are true, but it is not the case that $0 \leq 1 \leq \frac{1}{2}$.

You need to encode this quality that $s$ is the least such upper bound, that is for every upper bound $b$, the inequality you have is satisfied. In other words, for every $a \in A$ and every $b$ such that $b \geq c$ for all $c \in A$, $a \leq s \leq b$. Let me know if this helps.

tiral
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  • Thank you! This is very helpful. If I am following, the formalization would insted be something like: $$\forall a \in A : \forall b : b \ge c \ \forall c \in A : a \le s \le b$$ but maybe not $$\forall a,c \in A : \forall b : b \ge c : a \le s \le b$$ – isaac Jan 12 '24 at 02:16
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    It seems like this boiled down to order of universal quantification over the set, so I think this is a relevant post too: https://math.stackexchange.com/a/1882463/1098426 – isaac Jan 12 '24 at 02:23
  • I'm sorry, one more question. Via the linked post on quantification would the following be correct:

    $$ \forall a, c \in A : \forall b , (b \geq c \rightarrow a \leq s \leq b)$$

     – isaac Jan 12 '24 at 02:35
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    That is not correct. Again use $A = [0, 1]$ so that $s = 1$. Take, for example, $a = 0, c = \frac{1}{3}, b = \frac{2}{3}$. $a, c \in A$ and $b \geq c$, but $\frac{2}{3} = b < s = 1$ – tiral Jan 12 '24 at 02:40
  • Ah, that makes sense, I will come up with my own counterexamples if I have any other variations. Otherwise, I will stick with the initial notation we settled on. Thank you again. – isaac Jan 12 '24 at 02:42
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    Is the difficulty in translating the english "$b$ such that, for all $c$ in $A$, $b \geq c$" into a statement about quantifiers (aka you recognize that we need to some how force $b$ to be an upper bound in our definition and are just working out the details of how to concisely do this?) – tiral Jan 12 '24 at 02:43
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    No, I believe your second counter example showed where my line of thinking was flawed for how arbitrary elements in $A$ restrict the behavior of $\sup A$ and upper bounds for $A$. I initially just needed to see $\sup A$ at the same level of abstraction as upper bounds, even though the definition of $\sup A$ depends on the definition of upper bounds (being "least" upper bound). I do appreciate your help. – isaac Jan 12 '24 at 02:48