I have done everything else in my Modern Analysis class, but I wanted some assurances for the following greatest lower bounds/least upper bounds
a) $(0,3)$
I know the upper bounds would be 3 and the lower bound would be 0 but would the wording "least upper" and "greatest lower" make that non-existent?
c) $\{1, 1+\frac12, 1+\frac12+\frac14, \ldots \}$
I said 1 for the greatest lower bound
d) $\{\frac21, \frac{2\cdot 2}{1\cdot 3}, \frac{2\cdot2\cdot4}{1\cdot3\cdot3}, \frac{2\cdot2\cdot4\cdot4}{1\cdot3\cdot3\cdot5}, \ldots \}$
Would the lower not be 2, and would there be anything larger?
Please let me know if I am confusing terminologies or if I have made any mistakes. thanks!
"Least upper bound" does not mean the same thing as "maximum". Take for example, the set $(0,1)$, which has no maximum (in the sense that there is not a largest number which is less than one). An upper bound for a set, in general, is a number larger than (or equal to) every element of the set. So, $1$ is an upper bound for $(0,1)$, but so is $2$, or $\pi$, or $\sqrt{57}$. Each of these numbers is bigger than every number less than one, and so they are each an upper bound for the set $(0,1)$.
However, among the collection of upper bounds for the given set, there is a least such number. There is a number which is an upper bound for $(0,1)$ (i.e., not smaller than any number in $(0,1)$), and for which no smaller number can be an upper bound. That number is, in this case, $1$. The proof is short: any number in $(0,1)$ is less than one, by definition, hence $1$ is an upper bound. Conversely, for any number $c < 1$, you can find a number between $1$ and $c$ which belongs to $(0,1)$. For example, the number $\max\big(0.1, \frac{c + 1}{2}\big)$. This means that $c$ is not an upper bound for $(0,1)$, because it is not larger than (or equal to) every element in $(0,1)$.
On the flip side, if a set does indeed have a maximum element, then that element is automatically the least upper bound (also called supremum or just sup). For example, the set $(0,1]$ has a maximum element of $1$ and this is therefore the least upper bound. In another case, the set $\{1,2,3,\ldots\}$ has a minimum element of $1$ and this is therefore the greatest lower bound.
By the way, the definition of 'greatest lower bound' is the same, but flipped, as the least upper bound. That is, there may be many lower bounds for a given set, but there is one that is greater than them all (in size). This is the greatest lower bound, or infimum (or just inf).
With these in mind, I think you can find the answer to your examples. Your (a) is correct and your GLB for (c) is correct (where did (b) go?), and for the LUB notice that the elements of that set form an increasing sequence with some limiting value.
As for (d), this one is a bit trickier. I think if you plot the terms of that sequence, you will find that they alternate above and below some limiting value, and therefore the limit would be between the first and second terms (?)