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Trying to see if I understand the definition of bounded. Are these true:

  1. $(-1, 1)$ is bdd above by $\mathbb{R}$ (and below)
  2. $(-1, 1)$ is bdd above by $(-1, 1]\, $, and $1$ is the supremum of $(-1,1)$
  3. $[-1, 1]$ is bdd above and below by itself
  4. $(-1, 1)$ is not bdd above by $[-1,1)$
  5. $\mathbb{R}^+$ is not bdd above by $\mathbb{R}$ but it is bounded below, with an infimum of $0$.

Any mistakes or other elementary cases I should consider?

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    $\begingroup$ "Bounded" (in a metric space) typically means that the set is contained in a ball of finite radius. Bounded above (or below) in $\mathbb{R}$ typically means that there is some real number that is larger than (or smaller than, resp) any element of the set. When you say "bounded by", do you perhaps mean "properly contained in"? $\endgroup$
    – Xander Henderson
    Sep 2, 2017 at 18:17
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    $\begingroup$ In $\mathbb R$, a set cannot be bounded by another set; it can only be bounded by some real number. $\endgroup$ Sep 2, 2017 at 18:19
  • $\begingroup$ So the first bullet, for example, should be written "... is bdd above"? That is, given that the set in question is a subset of some other set in $\mathbb{R}$. $\endgroup$
    – Zduff
    Sep 2, 2017 at 18:28
  • $\begingroup$ Zduff: You are the expert on what you are asking, but the supremum of $(-1,1)$ in $\mathbb R$ is $1$. $\endgroup$
    – hardmath
    Sep 2, 2017 at 18:31
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    $\begingroup$ I've numbered your bullet points to make it easier to refer to them (I hope you don't mind). All of the intervals in 1--4 are bounded above by 1 (or any number larger than 1, and are bounded below by $-1$ (or any number less than $-1$. The question of proper containment is different. Hence my original comment. $\endgroup$
    – Xander Henderson
    Sep 2, 2017 at 18:31

4 Answers 4

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I have up-voted the answer by Xander Henderson. But since the subject line of the question says "Is every proper subset of the real numbers bounded by the set of real numbers?" it seems like a good idea to add that the set $\mathbb Z$ of all integers is a proper subset of $\mathbb R$ that is not bounded either above or below within $\mathbb R.$ (And I wrote "within $\mathbb R,$" not "by $\mathbb R.$")

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Each of the intervals in (1)--(4) is bounded in $\mathbb{R}$ (not by $\mathbb{R}$, but in). In each case, the given interval has infimum $-1$ (and is therefore bounded below in $\mathbb{R}$ by $-1$ and anything smaller), and supremum $1$ (and is therefore bounded above in $\mathbb{R}$ by $1$ and anything larger). (5) is slightly different: $\mathbb{R}^{+}$ is bounded below in $\mathbb{R}$ by 0 (this is the infimum), but unbounded above in $\mathbb{R}$.

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No proper subset of the real numbers is bounded by the set of real numbers, nor is every proper subset of the real numbers bounded in the real numbers (e.g., your fifth example). All of your examples are wrong, because you have fundamentally misunderstood the concepts of lower and upper bounds. Sets do not bound other sets; numbers bound sets. Let me try to clear things up for you.

In general, if $(S, \leq)$ is a poset and $T \subset S$, then an upper bound for $T$ is an element $s \in S$ such that $s \geq t$ for all $t \in T$. The plain English translation of this is that an upper bound of a set is an element of the ambient space that is greater than or equal to every element of said set. Similarly, a lower bound is less than or equal to every element of the set (just reverse the inequality).

You do not need the generality of a partially ordered set or metric space, however, so let me define upper bounds again, this time specifically for usual one-dimensional Euclidean space $\mathbb{R}$. Let $S$ be a subset of $\mathbb{R}$, and let $T$ be a subset of $S$. The number $s \in S$ is called an upper bound of $T$ in $S$ if and only if $s \geq t$ for all $t \in T$. Again, reversing the inequality gives the definition of a lower bound for $T$ in $S$.

Using one of your examples, $(-1,1)$ is not "bdd above by $\mathbb{R}$," because that does not make sense; however, $(-1,1)$ is bounded above by $1$, or $2$, or $100$, or indeed any real $x \geq 1$. Indeed, all four of your first examples are bounded above by $1$ and bounded below by $-1$.

Your fifth example is $\mathbb{R}_{>0}$, which is bounded below by $0$, but which is not bounded above in $\mathbb{R}$, because there is no real number $x$ such that $x>y$ for all $y \in \mathbb{R}_{>0}$. (Suppose there were such an $x$, then $y=x+1 \in \mathbb{R}_{>0}$ is bigger than it, which contradicts the assumption.)

You also mention infimums and supremums for 2 and 5, so let me define those, too.

Let $S$ be a subset of $\mathbb{R}$, and let $T$ be a subset of $S$. Then, the infimum of $T$, denoted $\inf T$, is the greatest least bound of $T$, i.e., it is the largest among $T$'s lower bounds. Similarly, the supremum of $T$, denoted $\sup T$, is the least upper bound of $T$, i.e., it is the smallest among $T$'s upper bounds.

In each of 1-5, the upper and lower bounds are also the supremums and infimums, respectively.

You should also look up the least-upper-bound property of $\mathbb{R}$, understand the difference between infimums and minimums as well as supremums and maximums, try the (routine) exercises below, and do some reading from your textbook, assuming you are using one.

Exercises:

  1. Find the upper and lower bounds of $\mathbb{R}_{\leq 0}$ in $\mathbb{R}$, should they exist.
  2. Prove that if $\inf T$ (respectively, $\sup T$) exists, then it is unique.
  3. Find $\sup \{x \in \mathbb{Q}: x^2<2\}$ and $\inf \{(-1)^n: n=1,2,3, \ldots \}$.
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That is why the real number system is 'Complete'. In case of open interval, say (a,b) we can take the deleted neighbourhood of a and b. Each member of the neighbourhood system lies in the set R. But, we cannot 'fix' a upper bound or lower bound. Sorry for giving a rather informal answer, but I hope it'll be of use.

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  • $\begingroup$ What you intended to do with those deleted neighborhoods is not stated in your answer and quite cryptic. But there are indeed upper bounds of the interval $(a,b)$ -- for example $b+1$ is such an upper bound, and $b+\frac 1 2$ is another, and $b$ is another. And there is a least upper bound, and that is $b.$ So what do you mean by saying "we cannot 'fix' a [sic] upper bound"? $\qquad$ $\endgroup$ Sep 2, 2017 at 19:26
  • $\begingroup$ @MichaelHardy i was talking about lub and glb s. $\endgroup$ Sep 2, 2017 at 19:30
  • $\begingroup$ I can't say that makes anything clearer. Certainly a bounded open interval $(a,b)$ has a greatest lower bound and a least upper bound. What is meant by saying we cannot "fix" those? $\endgroup$ Sep 2, 2017 at 19:31

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