MHB Maximal Elements in a Bounded Set

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The discussion focuses on the concept of maximal elements in a bounded set, specifically addressing questions related to a set defined by the expression {1 - 1/(n+1) | n ∈ ℕ}. It highlights that while the set S = {0, 1/2, 2/3, 3/4, ...} is bounded above by 1, 1 itself is not an element of S, making it the least upper bound. Consequently, no element in S can be considered maximal, as each element has a greater successor within the set. The conversation emphasizes the importance of understanding bounded sets and the definitions of upper bounds and maximal elements. Overall, the discussion clarifies misconceptions about maximal elements in relation to bounded sets.
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Could someone please check these questions? Please correct them if necessary, with an explanation if you could.
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Hi ertagon2,

Everything looks OK, except 5b. Think about the set $\displaystyle\left\{1-\frac{1}{n+1}\:\bigg|\: n\in\mathbb{N}\right\}$.
 
I agree with everything except 5.2. Consider an open interval.
 
castor28 said:
Hi ertagon2,

Everything looks OK, except 5b. Think about the set $\displaystyle\left\{1-\frac{1}{n+1}\:\bigg|\: n\in\mathbb{N}\right\}$.

I don't think I understand. Can you elaborate?
 
ertagon2 said:
I don't think I understand. Can you elaborate?
Hi ertagon2,

This is the set $\displaystyle S=\left\{0,\frac12,\frac23,\frac34,\ldots\right\}\subset\mathbb{Q}$. This set is bounded above (by $1$). In fact, $1$ is the least upper bound of $S$, but it is not an element of $S$.

No element of $S$ can be maximal, because, for each element $\left(1 - \dfrac{1}{n+1}\right)\in S$, $\left(1 - \dfrac{1}{n+2}\right)$ is greater and also an element of $S$.
 

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