Is $\lim\limits_{n\rightarrow\infty} n$ comparable to $\aleph_0$?

Solution 1:

The reason the answer is negative is that $$\huge\underline{\underline{\color{red}{\textbf{Cardinals are not real numbers.}}}}$$

What do I mean by that? For finite cardinals we can nicely match the natural numbers with the ordinals, the finite cardinals, the iterated sums of the unity of the real numbers, or the rationals, or the complex numbers, or whatever.

But once infinitary operations are involved (via limits or otherwise) we are no longer playing by the same rules.

It is true that $\lim_{n\to\omega}n=\aleph_0$ if you consider this sequence as a sequence of cardinals. But using $\infty$ means that you clearly don't think about these as cardinals, but rather as real numbers or something related. And these are two entirely distinct systems. The role of $\infty$ in analysis is entirely different than the role of $\aleph_0$ as a cardinal, or $\omega$ as an ordinal.

The above mixing that finite cardinals allow is to do between these systems is whence all these mistakes come from. And you're not alone in making them. Many people do, which is why I usually write the above line in huge letters, with several underlines, when I teach this stuff to my students. I want it to be comically rememberable to them, so they never again make this mistake.

On a side note, $2^{\aleph_0}$ and $\aleph_1$ are two distinct cardinals with two distinct definitions. Positing their equality is known as the continuum hypothesis, which the standard axioms of set theory can neither prove nor disprove.

Solution 2:

There is an ordered set of extended natural numbers $\mathbb{N} \cup \{ \infty \}$.

The ordered class of cardinal numbers has an initial segment $\mathbb{N} \cup \{ \aleph_0 \}$.

These two ordered sets happen to be isomorphic. This fact is pretty much the entirety of the relationship between $\infty$ and $\aleph_0$.

However, there is something else along these lines that may be interesting. If you consider the hyperreal numbers of nonstandard analysis, the hyperreals contain a lot of infinite numbers $H$. However, every hyperreal (including the infinite ones) satisfies $-\infty < H < \infty$.

Solution 3:

To summarize information I've gotten from the existing answers and discussion in comments:

• Cardinals and real numbers are not comparable with the standard relations for real numbers nor those for cardinals (e.g. the usual $=$, $<$, etc., $<$ for real numbers is not the same $<$ as for cardinals, etc.).
• $\infty$ is not a cardinal either and so isn't comparable to cardinals
• One could probably define any arbitrary relation they want between $\aleph_0$ and $\infty$ and it would be of no consequence to mathematics.

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Now, how about the following:

Let $\infty$ represent the length of the real line. We have that $1=\mu\left([n,n+1]\right)$ the length of each segment between consecutive integers for $\mu$ the standard length measure.

Thus the length of the real line is $\infty=\displaystyle\sum_{i\in\mathbb Z}1$.

Since there are exactly $\aleph_0$ unit length intervals for consecutive integers (and exactly $\aleph_0$ consecutive intervals of any finite length, of course), then to get the length of the real line, we just count these unit intervals, hence the length of the real line would be $\aleph_0$ if we were to allow $\aleph_0$ to represent a spatial magnitude.

So the only reasonable/natural comparison would be $\infty=\aleph_0$ if one were to make a comparison. NOTE: The $=$ used here is not the equals sign used to show identity of real numbers! Nor is it the equals sign used to equate cardinals!