Least upper bounds and greatest lower bounds: open vs. closed intervals

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I tried looking through other questions but couldn't find this one answered. I'm working through Lax and Terrell's 'Calculus with Applications'.

The book asks to state whether the following intervals have GLBs and LUBs and if so, what they are: $(8,10)$, $(8,10]$.

My initial thought was that an open interval can't have a LUB or GLB as they don't include the end points and thus there is always a number arbitrarily closer to the end point that doesn't quite 'reach' it. However, I've also seen some discussion such that, e.g., the LUB of both half-open and open intervals are the same given that $9.99999... = 10$.

Help appreciated here as I want to understand this before proceeding!

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Every bounded non-empty subset of $\Bbb R$ has a greatest lower bound and a lowest upper bound. In the case of $(8,10)$ and of $(8,10]$, these are $8$ and $10$ respectively. The fact that $10$ belongs to the second interval but not to the first one changes nothing. And the fact that $9,99999999\ldots=10$ also changes nothing.

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IF the lub of a set is in the set then it is the maximum of the set. IF the glb of a set is in the set then it is the minimum of the set.

(0, 1), (0, 1], [0, 1), and [0, 1] all have glb 0 and lub 1.

(0, 1) has neither a minimum nor a maximum.

(0, 1] has no minimum but has maximum 1.

[0, 1) has no maximum but has minimum 0.

[0, 1] has minimum 0 and maximum 1.

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Whether or not $(8,10)$ has a l.u.b. depends on what space/set you're working with to begin with, i.e. what the "parent space" is.

If you're working in $X = (8,10)$ then the set $(8,10)$ has no l.u.b. in $X$.

Another example is the set $A =$ {$x^2 < 2: x \in \mathbb{Q}$} has no l.u.b. in the parent set $\mathbb{Q}$. However, if the parent set was $\mathbb{R}$, then of course the l.u.b of $A$ would be $\sqrt{2}$.

So we could conclude that a set $S \subseteq T$ has a l.u.b. in the set $T\ $ if:

$\ T$ is complete and $\exists s \in T$ which is an upper bound of the set $S$.