Are there fewer reals on $(0, 1)$ than on $(1,\infty)$?

Question

I know that the cardinality of the sets of real numbers $(0, 1)$ and $(1, \infty)$ are equal.

So what is the fallacy in this argument?

For every real on $(0, 1)$, we can add any integer $n$ to it and get a number on $(1, \infty)$, with the number from the first set as the fractional part of our new number. However, this can be applied to every real on $(0, 1)$, with every integer greater than one... seeming to suggest a larger cardinality of the second set than the first.

Answer 1

Your argument shows that the cardinality of $(1,\infty)$ equals the cardinality of $(0,1)$ times the cardinality of the positive integers. In other words,

$$\mathfrak c=\mathfrak c\cdot\aleph_0$$

That is also known in other ways. The equation comes from the definition of the product of cardinal numbers.

The source of the "paradox" is that an infinite set can have the same cardinality as a proper subset. This cannot hold for finite sets but often does for infinite sets. The usual way to introduce that "paradox" is to show that the sets $\{1,2,3,\ldots\}$ and $\{0,1,2,3,\ldots\}$ have the same number of elements, though the second set has "one more" element than the first. For a dramatization of this, see Hilbert's Hotel.

Answer 2

The problem is you do not have one mapping, you have a family of maps indexed by $\mathbb{N}$. In order to deduce something formally about cardinality, one must talk about a single map, not a bunch of maps.

Also, each of your maps embeds $(0,1)$ into $(1,\infty)$, which only shows that the cardinality of $(0,1)$ is less than or equal to the cardinality of $(1,\infty)$. It does not conclude anything about strict inequality of their cardinalities, for this would require a proof stating the non-existence of a bijection between your sets.

Answer 3

Your argument does not show a larger cardinality for $(1,\infty)$. You have shown that $\mathfrak {c=c}\cdot \aleph_0$, which is true

Answer 4

This doesn't explicitly answer your question, but may contribute to your understanding.

What you have observed (essentially) is that there is a proper subset of the real numbers with the same cardinality of the whole set. Here "cardinality" means "matches using a one to one and onto function". The next step is to realize that's not a contradiction - it's the definition of what it means for a set to be infinite.

The usual first instance of this "contradiction" (or paradox) is that there are the same number of even integers as integers.

The other answers here are correct, but perhaps more technical than what you need.

Answer 5

The essential mistake here is the assumption that if you can establish a One-To-Many mapping from A to B, then you have proven that Cardinality(A) < Cardinality(B), or just in general, that they are not equal. While this happens to be true for finite sets, this most definitely is not true for sets of transfinite cardinality.

Rather, you must rely on the definition of Equality for set-cardinality, which is that if you can establish any 1-to-1 mapping from A onto B then Cardinality(A) = Cardinality(B). Therefore, to establish non-equal cardinality, you must show that there is no 1-to-1 onto mapping possible between them (which is what the Diagonal Proof does for Natural numbers Vs. Real numbers).