It's important to be precise in your context, here. If you are discussing a poset $P$, and wondering if $P$ has a least upper bound in the poset $P$, then you are wondering precisely if $P$ has a maximum element.
However, if you were discussing a subset of $P$, say $A$, and wondering if $A$ has a least upper bound in the poset $P$, then that's a different matter. If $A$ has a maximum element, then it will readily be the least upper bound of $A$. However, if $A$ has no maximum element, but only maximal elements--such as $A=\{a,b\}$ with $a,b$ incomparable--then the least upper bound of $A$ (if it exists) will necessarily be a member of $P$ that is not in $A$.
In your particular example with $A=\{4,5,7\}$, note that $4$ is incomparable with both $5$ and $7$, and that $5$ is greater than $7$. Hence, any upper bound of $A$ need only be an upper bound of $4$ and $5$, and any lower bound of $A$ need only be a lower bound of $4$ and $7$. The set of all upper bounds of $A$ is $\{1,2,3\}$ and this set has a least element, namely $3$, so $\sup A=3$. The only lower bound of $A$ is $8$, so $\inf A=8$.
Note that in general, $\sup A$ (if it exists) is the minimum element of the set of upper bounds of $A$, and $\inf A$ (if it exists) is the maximum element of the set of lower bounds of $A$. If $A$ has no upper (lower) bounds, then $\sup A$ ($\inf A$) doesn't exist. For example, if we didn't have the node $8$ in your example $W$, then $\inf A$ wouldn't exist. If the set of upper (lower) bounds of $A$ is non-empty, but has no minimum (maximum) element, then $\sup A$ ($\inf A$) doesn't exist. For example, suppose we were to add another node, $9$, to your example, on the same level as $8$, with $9\prec 6$, $9\prec 7$, and with $8$ and $9$ incomparable. In that case, the set of lower bounds of $A$ would be $\{8,9\}$, but since $8$ and $9$ are incomparable, that set has no maximum element, meaning $\inf A$ wouldn't exist.
As far as the real numbers are concerned, a set has an infimum if it is nonempty and bounded below, and a supremum if it is nonempty and bounded above. As for the work you've done:
a) looks good.
For b) it should say "...and $x\leq 1$" not "...and $1\leq x$". Other than that, looks good.
For c), as long as you do not include $0$ as a natural number, everything looks fine.
For d), yes, that is one way to represent the real numbers. So the set is unbounded on both ends.
For e), you are talking about a range of real numbers. The notations $(x,y),[x,y),(x,y]$ and $[x,y]$ denote intervals real numbers. If it were a set of integers, the set would be defined with curly brackets, $\{0,2\}$ represents the set containing only the two integers $0$ and $2$. You might also occasionally see the notation $[0,2]\cap \Bbb{Z}$, which would be equivalent to the set $\{0,1,2\}$. Hence, the two upper bounds to consider for this problem are $4$ or $5$.
Now that you have established an understanding of these five sets and picked out correct lower/upper bounds, can you identify which of those are supremums and infimums of your respective sets?
Best Answer
Your idea is exactly right, but we often have a convention that a set with no upper bound, such as the positive real numbers, has a supremum of "$\infty$", and a set with no lower bound has an infimum of "$-\infty$". In this sense every set has a supremum and an infimum, although it may not have a minimum or a maximum.