Is x < ∞ always finite? Exploring the concept of infinity and finiteness

[V0ODO0CH1LD]

If something is strictly less than something else, is it definitely finite?

Because if x < y, and the biggest thing y can be is ∞, then x < ∞, but does that mean x is finite? Couldn't x = (∞ - 1)? Isn't that still technically infinity? What if x = (∞ - ε)?

 

[HallsofIvy]

The classical "real number system" does NOT include "infinity" as a number and you cannot have an operation on real numbers, such as "$\infty - 1$" or "$\infty- \epsilon$", or equations involving "infinity".. As for inequalities, $x< \infty$ simply means that x can be any real number while $y< x< \infty$ says that x can be any real number larger than y.

There are several different ways of extending the real number system to include one or more "infinities" as well as "infinitesmals". What operations, equations, or inequalities you are allowed to use depends on the particular extension.

 

[V0ODO0CH1LD]

I came to that because I was trying to prove that $\mathcal{T}=\{U\subset X:X-U\ is\ finite\ or\ X\}$ is a topology on $X$.

Finite sets are defined as sets that have a bijection with a proper subset of the natural numbers (i.e. $S$ is finite if there is a bijection $S\rightarrowtail \{1,...,n\}$ for some n in the naturals).

The empty set and $X$ are in the topology since $X-X\rightarrowtail \emptyset$, and $X-\emptyset=X$.

But for $U_1\cup U_2\cup ...$ to be in $\mathcal{T}$, $X-(U_1\cup U_2\cup ...)$ has to be finite. Which means $(X-U_1)\cap (X-U_2)\cap ...$ has to be finite because $X-(U_1\cup U_2\cup ...)=(X-U_1)\cap (X-U_2)\cap ...$ from the elementary algebra of sets.

And if $(X-U_i)\rightarrowtail \{1,...,n_i\}$, I know that $(X-U_1)\cap (X-U_2)\cap...\rightarrowtail\{1,...,m\}:m<\sum n_i$, since $|S\cap T| < |S|+|T|$ if $S$ and $T$ are not empty.

But does the fact that $m<\sum n_i$ prove that arbitrary unions of sets in the topology are finite?

EDIT: because I could keep summing $n_i$'s until it got arbitrarily close to infinity. And although $m$ would still be less than that sum, I am not sure if it is finite.

 

[PeroK]

A subset of a finite set is finite. So, if X - U_i are finite, then the intersection of X - U_i is a subset of X - U_1 so must be finite.

By the way, your definition of finite appears to exclude the empty set.

 

[V0ODO0CH1LD]

A subset of a finite set is finite. So, if X - U_i are finite, then the intersection of X - U_i is a subset of X - U_1 so must be finite.

I was trying to prove that, but actually got it. I realized that $m$ in that case ins't just less than $\sum n_i$, it's actually less than or equal to all $n_i$, making it certainly less than infinity :)

By the way, your definition of finite appears to exclude the empty set.

Isn't the empty set a proper subset of the naturals? In that case it has a bijection with itself.

 

[R136a1]

Isn't the empty set a proper subset of the naturals? In that case it has a bijection with itself.

Your definition of finite is wrong, but not for the reason of the empty set. The set of even numbers $E = \{2,4,6,...\}$ is also a proper subset of the naturals, so it is trivially in bijection with a subset of the naturals. But it's not finite.

You also provide a second definition where something is in bijection with $\{1,...,n\}$. But the empty set would be excluded here. Since there is no $n$ such that $\{1,...,n\}$ is finite.

 

[V0ODO0CH1LD]

Your definition of finite is wrong, but not for the reason of the empty set. The set of even numbers $E = \{2,4,6,...\}$ is also a proper subset of the naturals, so it is trivially in bijection with a subset of the naturals. But it's not finite.

You also provide a second definition where something is in bijection with $\{1,...,n\}$. But the empty set would be excluded here. Since there is no $n$ such that $\{1,...,n\}$ is finite.

That's true, although $\{1,...,n\}$ was just an illustration. But what is one correct definition of finite sets? Could I say that a finite set is bijective to a set $S_n$, where $S_n$ is the set of all natural numbers (including zero) less than $n$? Then $\emptyset\rightarrowtail S_0$. Or do I have to make a special case for the empty set? Like just use the definition $\{1,...,n\}$ and separately state that the empty set is finite with cardinality zero.

 

[PeroK]

http://en.wikipedia.org/wiki/Finite_set

 

[1MileCrash]

I came to that because I was trying to prove that $\mathcal{T}=\{U\subset X:X-U\ is\ finite\ or\ X\}$ is a topology on $X$.

Finite sets are defined as sets that have a bijection with a proper subset of the natural numbers (i.e. $S$ is finite if there is a bijection $S\rightarrowtail \{1,...,n\}$ for some n in the naturals).

The empty set and $X$ are in the topology since $X-X\rightarrowtail \emptyset$, and $X-\emptyset=X$.

But for $U_1\cup U_2\cup ...$ to be in $\mathcal{T}$, $X-(U_1\cup U_2\cup ...)$ has to be finite. Which means $(X-U_1)\cap (X-U_2)\cap ...$ has to be finite because $X-(U_1\cup U_2\cup ...)=(X-U_1)\cap (X-U_2)\cap ...$ from the elementary algebra of sets.

And if $(X-U_i)\rightarrowtail \{1,...,n_i\}$, I know that $(X-U_1)\cap (X-U_2)\cap...\rightarrowtail\{1,...,m\}:m<\sum n_i$, since $|S\cap T| < |S|+|T|$ if $S$ and $T$ are not empty.

But does the fact that $m<\sum n_i$ prove that arbitrary unions of sets in the topology are finite?

EDIT: because I could keep summing $n_i$'s until it got arbitrarily close to infinity. And although $m$ would still be less than that sum, I am not sure if it is finite.

I had to prove that the same set was a topology, we called it the "cofinite" topology. We had it slightly different (without the "or X" and the whole thing unioned with the set of the empty set) but it is the same.

I don't think you need any of these ideas at all to prove that this collection is a topology. I think you only need to consider what happens to cardinality of unions and intersections. We can say that an intersection of any number of sets has at most the cardinality of the smallest set, and we can say that the union of two sets has at most a cardinality equal to the sum of the two cardinalities. Now, you know that the original cardinalities before unions/intersections in question are all finite, so then what can you conclude?

 

[1MileCrash]

EDIT: because I could keep summing $n_i$'s until it got arbitrarily close to infinity. And although $m$ would still be less than that sum, I am not sure if it is finite.

Remember that you only need to show that finite intersections are in the topology. Your first part is regarding unions, your second part is regarding intersections, so you can prove it by simply intersecting two sets.

 

[R136a1]

That's true, although $\{1,...,n\}$ was just an illustration. But what is one correct definition of finite sets? Could I say that a finite set is bijective to a set $S_n$, where $S_n$ is the set of all natural numbers (including zero) less than $n$? Then $\emptyset\rightarrowtail S_0$. Or do I have to make a special case for the empty set? Like just use the definition $\{1,...,n\}$ and separately state that the empty set is finite with cardinality zero.

Yes, both work!

 

[V0ODO0CH1LD]

Remember that you only need to show that finite intersections are in the topology. Your first part is regarding unions, your second part is regarding intersections, so you can prove it by simply intersecting two sets.

Not in that case, I turned the problem of proving arbitrary unions are in the set into proving that arbitrary intersections of finite sets are finite. I didn't even mention the finite intersection property of topology in that post.

 

[R136a1]

Not in that case, I turned the problem of proving arbitrary unions are in the set into proving that arbitrary intersections of finite sets are finite. I didn't even mention the finite intersection property of topology in that post.

So, take $A_i$ finite for $i\in I$ and $I$ nonempty. Take some $j\in I$, then

$$\bigcap_{i\in I} A_i\subseteq A_j$$

and $A_j$ is finite by hypothesis. Can you deduce the intersection is finite too?

 

[1MileCrash]

Not in that case, I turned the problem of proving arbitrary unions are in the set into proving that arbitrary intersections of finite sets are finite. I didn't even mention the finite intersection property of topology in that post.

Yes, arbitrary intersections of finite sets are finite, which we consider from arbitrary unions of members of the topology.

And, finite unions of finite sets are finite, which we consider from finite intersections of members of the topology.

So where is the problem?

 

[V0ODO0CH1LD]

Yes, arbitrary intersections of finite sets are finite, which we consider from arbitrary unions of members of the topology.

And, finite unions of finite sets are finite, which we consider from finite intersections of members of the topology.

So where is the problem?

Nowhere, I've just never proved that to myself. That was the whole point.. but now I did!

 

[1MileCrash]

Nowhere, I've just never proved that to myself. That was the whole point.. but now I did!

Well, it's just the smallest of a possibly infinite group of finite numbers, is finite. And the sum of finitely many finite numbers, is finite.