Example of a set which is not bounded?

5,667

Solution 1

How about $a_n = (-1)^n n$ for an oscillating unbounded sequence.

Solution 2

Within the set $\mathbb{R}\cup\{\infty\}\cup\{-\infty\}$, the set $\mathbb{R}$ is bounded, but usually what is meant by "bounded" is having upper and lower bounds within $\mathbb{R}$. By that usual definition $\mathbb{R}$ is not bounded. Among other unbounded sets are the set of all natural numbers, the set of all rational numbers, the set of all integers, the set of all Fibonacci numbers. All finite sets are bounded. Many infinite sets are bounded as well---for example, the set of all numbers between $0$ and $1$, and the numbers in the sequence $1,1/2, 1/3, 1/4,\ldots\ {}$.

5,667

Chibueze Opata

Updated on December 06, 2022

Comments

Chibueze Opata 6 months

It suddenly occurred to me that the set of real numbers is bounded. So suddenly, I'm wondering: What is an example of a set that is unbounded.

NOTE: This question was triggered when I came across oscillating sequences (which can be finite or infinite), so I was wondering an example of an infinite oscillating sequence.

Thanks.
- M Turgeon about 11 years
  
  The subset of rational numbers (the real numbers are not bounded)
- Chibueze Opata about 11 years
  
  Really? I don't think so.
- Martin Argerami about 11 years
  
  If you think the rationals or the naturals are bounded, then give us an upper bound.
- Chibueze Opata about 11 years
  
  The way I learnt it is that it's bounded above by + infinity.
- Martin Argerami about 11 years
  
  Yes, of course you can do that. But what you are doing is just taking your own definition of "bounded", which does not agree with the common use. Which in particular means that wherever you see the word "bounded" used by someone else, it doesn't mean what you think it means.
- Chibueze Opata about 11 years
  
  I already explained it's not my own definition. I guess the definition is misleading but Michael Hardy explained this issue.
- Martin Wanvik about 11 years
  
  @Chibueze Opata: I wouldn't call the definition misleading, I'd call it pointless - after all, every subset of the reals is bounded above by $+\infty$.
- Chibueze Opata about 11 years
  
  I'd like to say that into the Professor's face...I'll quote you as well :)
- Martin Argerami about 11 years
  
  @ChibuezeOpata: what definition are you talking about? You'll be hard pressed to find a mathematician that says that $\mathbb{N}$ or $\mathbb{R}$ is bounded. The word "bounded" is very common in the mathematical literature, and it does not mean what you say it means. According to you, every function is "bounded", every subset of $\mathbb{R}$ is "bounded"... As Martin Wanvik said above, what's the point of even using the word bounded then?
Chibueze Opata about 11 years

:) Can you provide example, say starting from -1 for this set?
Ross Millikan about 11 years

@ChibuezeOpata: This is $-1,2,-3,4,-5,6 \ldots $
Chibueze Opata about 11 years

This was really a useful answer. Thanks.
M. Vinay almost 9 years

@ChibuezeOpata Note that $\mathbb{R} \cup \{\infty, -\infty\}$ is not the set of real numbers. It is usually called the system of "extended reals": en.wikipedia.org/wiki/Extended_real_number_line
Chibueze Opata almost 9 years

Nice, what I had been looking for actually is a defined sequence that can take up to negative and positive infinities... Which is what Matt provided.