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I tried to find the answer to my questions for a while but did not succeed, and I hope that was not only because of deficits in my search terms.

My question is as follows:

Let $\mathcal{A}$ be a set of 3-dimensional positive real numbers, i.e. $\mathcal{R}^{+} \times \mathcal{R}^{+} \times \mathcal{R}^{+}$. I am now looking for a term that describes a subset $\mathcal{B}$ of $\mathcal{A}$ that may include points of $\mathcal{A}$ but not $\infty$.

The background is that $\int_{\mathcal{B}}1\;dx$ can be interpreted as the (geometric) volume of $\mathcal{B}$ if $\mathcal{B}$ does not include $\infty$.

Is this a closed subset, or a proper subset? As far as I understood both are not correct. I would like to says something like

"Define $\mathcal{B}$ as a ... subset of $\mathcal{A}$"

and this expression shall make clear $\mathcal{B}$ may contain any points in $\mathcal{A}$ but may not include infinity. Or is there no specific expression and shall simply say

"Define $\mathcal{B}$ as a subset of $\mathcal{A}$, where non of its limits is infinity" ?

Thanks for every suggestion, Johannes

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2 Answers 2

You are slightly confused about notation in your question, but I think I understand what you are asking.

You say that $B\subset A$ may not include infinity, but in reality, since $A$ only contains triplets of real numbers, even the set $A$ does not include infinity. This means that for any set $B\subseteq A$, the statement "$B$ does not include infinity" is perfectly correct.


That said, what I think you are asking is how to describe a set that does not contain arbitrarily large elements. For example, $S=\{(x,x,x)|x>0\}$ contains the numbers $$(1,1,1),(2,2,2),\dots, (1000,1000,1000),\dots, (N,N,N),\dots$$ where $N$ is arbitrarily large. If that is indeed what you are asking, then your terms "closed" and "proper" are incorrect:

  • A set is closed intuitively, if it contains its own border. This means that $A$ itself is a closed set and, since $A$ obviously contains arbitrarily large numbers, this is not the term you need. Also, the set $S$ described above is also closed.

  • A set is a proper subset of $A$ if it is not equal to $A$. This means that $A\setminus\{(1,1,1)\}$ is a proper subset of $A$, but again, it contains arbitrarily large numbers, so "proper" is not the term you need. Again, the set $S$ described above is also closed.

    and the term you are looking for is BOUNDED

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Thanks a lot for your explanation! However may I ask you to support my understanding and explain to my why "for any set B⊆A , the statement "B does not include infinity" is perfectly correct."? Since A includes the point $\infty \times \infty \times \infty$, this could also be in B, which would then include infinity - right? –  Joe1979 Jul 31 at 8:03
1  
@Joe1979 No, $A$ does not include $(\infty,\infty,\infty)$. $A$ includes only triples $(x,y,z)$ where $x,y,z$ are positive real numbers, and, since $\infty$ is not an element of $\mathbb R$, $(\infty,\infty,\infty)$ is also not an element of $A$ –  5xum Jul 31 at 8:13
    
OK, thanks for the clarification. –  Joe1979 Jul 31 at 9:16

I think you are describing a bounded subset.

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Thanks, I will use bounded now! –  Joe1979 Jul 31 at 8:07

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