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I'm reading a book on Real Analysis, which has the following definition of least upper bound:

Definition of least upper bound

The paragraph after the bullet points leads me to understand that the least upper bound of a subset of the reals is necessarily outside the set. The paragraph after that leads me to understand the least upper bound is inside the set.

Is it necessarily one or the other?

I'm thinking open sets like $(0,1)$ have a least upper bound of one, which is outside the set, and sets with closed bounds like $(0,1]$ have their least upper bound inside the set.

Am I understand understanding the concept of least upper bounds correctly, at least so far?

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    $\begingroup$ Yes, it can be inside or outside the set. If you don't know, write $\sup S$. If you know it is inside, write $\max S$. $\endgroup$
    – max_zorn
    Commented Jan 22, 2018 at 8:24
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    $\begingroup$ "The paragraph after the bullet points leads me to understand that the least upper bound of a subset of the reals is necessarily outside the set." It definitely does not mean that, study it again until your understanding changes. $\endgroup$
    – Ben Voigt
    Commented Jan 22, 2018 at 15:03
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    $\begingroup$ No. "It is not necessarily the case that X" is not equivalent to "it is necessarily the case that ~X". $\endgroup$
    – Daniel R. Collins
    Commented Jan 22, 2018 at 15:43
  • $\begingroup$ Hello, can you tell me what book it is? It feels easy to read. $\endgroup$
    – alu
    Commented Apr 9, 2021 at 16:38

2 Answers 2

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It need not be inside or outside as you have illustrated in your example.

In the last paragraph, notice a few keywords of which I will highlight.A non-empty finite subset has a greatest element, in this case, they must be inside the set.

If a set has a greatest element, then it must be inside the set.

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  • $\begingroup$ Thank you! So, I'm familiar with the idea that the real numbers are uncountably infinite. My understanding is that any continuous, open subset of the reals with ordinality greater than one is also uncountably infinite. So, its least upper bound must be outside the set, correct? $\endgroup$
    – StudentsTea
    Commented Jan 22, 2018 at 9:02
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    $\begingroup$ Do you mean set of the form of $[a, b), (a,b), (a, \infty)$ and $(-\infty, \infty)$. The first two have least upper bound that are outside the set. The last $2$ do not have a least upper bound. $\endgroup$
    – Siong Thye Goh
    Commented Jan 22, 2018 at 9:07
  • $\begingroup$ Ah! Good point. I was only considering sets like the first two. Hadn't considered sets defined using infinity. $\endgroup$
    – StudentsTea
    Commented Jan 22, 2018 at 9:39
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Your guess of $\sup(0,1)$ and $\sup (0,1]$ is correct. Specifically, you can prove that $\sup S\in S$ if and only if $\max S$ exists (id est, an element of $S$ which is larger or equal to any other element of $S$), in which case $\sup S=\max S$.

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