# Is the least upper bound of a set necessarily outside the set?

I'm reading a book on Real Analysis, which has the following 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?

• Yes, it can be inside or outside the set. If you don't know, write $\sup S$. If you know it is inside, write $\max S$. Commented Jan 22, 2018 at 8:24
• "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. Commented Jan 22, 2018 at 15:03
• No. "It is not necessarily the case that X" is not equivalent to "it is necessarily the case that ~X". Commented Jan 22, 2018 at 15:43
• Hello, can you tell me what book it is? It feels easy to read.
– alu
Commented Apr 9, 2021 at 16:38

• 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. Commented Jan 22, 2018 at 9:07
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 $\sup S=\max S$.