Finding the supremum and infimum of the following set

Question

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 $\sup(S \cup \{u\}) = \sup \{s,u\}$.

Thanks in advanced.

Answer 1

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

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

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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.

Answer 2

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.