[Math] Does this set $\{x + 1 : x \in \mathbb{R}\}$ have an upper and lower bound

elementary-set-theory

I understood the concepts of maximum, minimum, lower and upper bound, supremum and infimum, but I am still not so confident when it comes to say if a set has those or not.

In one of the exercises I have to do, it's asking us to say if the sets are bounded from below or above, and to give 2 upper and 2 lower bounds if they exits.

I will proceed through all the sets, and I will try to give my answer, so that you can tell me where I am wrong.


a) $\{ 1, 4, 9, 16, 25\}$

This set is bounded from below and from above.
Two examples of lower bound are: $1$ and $0$, since for all $x$ in the set, $1\leq x$ and $0\leq x$. Two examples of upper bound are: $25$ and $26$, since for all $x$ in the set, $25 \geq x$ and $26\geq x$.

b) $\{ x \in \mathbb{R} : x < 0\}$

This set is bounded from above, but not from below, since it goes to $-\infty$. Two examples of upper bounds are $0$ and $1$, since for any number $x$ in the set, $x \leq 0$ and $x \leq 1$.

The range of this set can be represented by:

$$\left(-\infty, 0\right)$$

c) $\{1 + \frac{1}{2^n} : n \in \mathbb{N}\}$

This set is bounded from below and above. Two examples of upper bound are $\frac{3}{2}$ or any other number greater than $\frac{3}{2}$, since for any number $x$ in the set $x \leq \frac{3}{2}$. Two examples of lower bound are $1$ or any other number less than $1$, since for any number $x$ in the set, $1 \leq x$.

Basically, the set can be represented with the following range:

$$\left(1, \frac{3}{2}\right]$$

d) $\{x + 1 : x \in \mathbb{R}\}$

This set has no upper bound or lower bound, if we do not consider $-\infty$ or $\infty$ as respectively the lower and upper bound. This set basically represents the real numbers, right?

e) $[0, 2] \cup (3, 4)$

In this case, I have a doubt. Are we talking about ranges of real numbers or natural numbers? Because if we are talking about natural numbers, then the upper bounds could be $3$ and $4$, since they $\not\in [0, 2] \cup (3, 4)$. Otherwise the 2 upper bounds could be $4$ or $5$.

Two lower bounds are for example $0$ or $-1$.

Best Answer

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?