# Finding the supremum and infimum of the following set

Let $$A = (2,5) \cup \{7\}$$. Find the supremum and infimum of $$A$$.

Firstly, I claim that $$\sup A = 7$$ and $$\inf A = 2$$. But, I got confused that in $$A$$, there is a singleton set. How to approach the supremum? And also, the infimum?

An approach that I know is using this theorem:

Let $$S \subseteq \Bbb R$$ and suppose that $$s:= \sup S$$ belongs to $$S$$. If $$u \notin S$$, then $$\sup(S \cup \{u\}) = \sup \{s,u\}$$.

• I think you can use the above theorem. For another approach, I'm still have no idea. Aug 24, 2021 at 3:54
• Do you mean $A=(2,5)\cup\{7\}$ or $A=\{2,5\}\cup\{7\}$?
– Joe
Aug 24, 2021 at 10:23
• @Joe $A=(2,5) \cup \{7\}$. Aug 25, 2021 at 9:03

A sequence that converges to the supremum is $$7, 7, \ldots$$.

$$7$$ can be taken since it is an element of $$A$$.

Edit:

$$7$$ is an upper bound of $$A$$. Furthermore, any upper bound of $$A$$ must be bigger than $$7$$, hence $$7$$ is the least upper bound.

Similarly for the lowerbound.

• Then? How to show the supremum? Aug 24, 2021 at 7:50
• What do you mean about 7 can be taken ? Aug 24, 2021 at 7:52
• When he means how to approach the supremum, is he asking how to construct a sequence in $A$ such that it converges to $7$? that's how I interpret his question. I think that's the source of confusion since he is highlighting the singleton part. Aug 24, 2021 at 7:56

Just remember (or look) the definitions of supremum(least upper bound) and infimum(greatest lower bound). Then you can see that $$\sup A=7$$ and $$\inf A=2$$.

Let's look at the supremum case as the name suggest it must be the least upper bound so is $$6$$ an upper bound for $$A$$?. No because we have $$\{7\}$$ in our set. You can continue like this.

• How to show the supremum? Aug 24, 2021 at 7:51