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$[0, 1] \approx (0,1) \approx \mathbb{R}$, for example.

Intuitively, it seems that the infinity of $\mathbb{R}$ is of a different nature than that of the intervals; with $\mathbb{R}$ I can “explode” towards infinity, whereas with the intervals, in a seemingly opposite fashion, I can “dig” infinitely before reaching the bounds. Here I envision an infinitely straight line on the $x$-axis, versus a line from 0 to 1 with a scale that gets infinitely dense at the ends, as if you mapped all values of $\text{arctan}(x + \frac{\pi}{2})$ to a point on the line.

My question then is, how are we jamming a supposedly unbounded line into somewhere that is bounded? How can I even “grab” the ends of this line if they do not exist? Does this have to do with the term “dense” (e.g. dense nowhere, etc.) that I seem to see everywhere?

I’ve just completed a first course in mathematical logic, and we concluded with cardinalities and a brief introduction to analysis (bounds and the Completeness Axiom). I would be satisfied with a conceptual explanation, however something canonical, even if it is beyond my level, is what I am hoping for. Any guidance towards learning about things of the same nature as this is also greatly appreciated, even the names of such subjects.

Asaf Karagila
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kusa
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    Here's another example that may help to show irrelevance of being bounded: $\mathbb Z$ (unbounded in both directions) and ${0,1}\cup {\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots} \cup {\frac{2}{3},\frac{3}{4},\frac{4}{5},\ldots}$ (bounded, and with both a max and min in the set) via the correspondence $0\rightarrow0,$ $1\rightarrow1,$ $-1\rightarrow\frac{1}{2},$ $2\rightarrow\frac{2}{3},$ $-2\rightarrow\frac{1}{3},$ $3\rightarrow\frac{3}{4},$ $-3\rightarrow\frac{1}{4},$ $4\rightarrow\frac{4}{5},$ $-4\rightarrow\frac{1}{5},$ $5\rightarrow\frac{5}{6},$ $-5\rightarrow\frac{1}{6},$ $\ldots$ – Dave L. Renfro Apr 25 '19 at 08:40

2 Answers2

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When we say $\mathbb R$ is unbounded, it is usually in the context of $\mathbb R$ as a metric space. You may not have encountered metric spaces yet. A metric on a space $X$ is a function $d: X \times X \to \mathbb R$ such that

  • $d(x, y) = d(y, x)$ for all $x, y$
  • $d(x, x) \geq 0$ for all $x$
  • $d(x, y) = 0$ if and only if $x = y$

Then $(X, d)$ is referred to as a metric space. $d$ is meant to capture the concept of "distance". On $\mathbb R$, the usual metric is defined by $d(x, y) = \lvert x - y \rvert$ — the obvious choice for the distance between two points. You can see why having a definition of distance is useful to define unboundedness.

A metric space $(X, d)$ is unbounded if we can find points that are arbitrarily far apart. Rigorously we require that for all $M > 0$, there exists $x, y \in X$ such that $d(x, y) > M$. You can see that this is true for $\mathbb R$ but not $[0, 1]$ or $(0, 1)$. I can't really see how denseness is relevant.

Now are there alternate metrics on $\mathbb R$ other than the usual metric which would make $\mathbb R$ bounded? Yes! Something like $d(x, y) = 1$ if $x \neq y$ and $d(x, y) = 0$ otherwise would work. Or $d(x, y) = \lvert f(x) - f(y) \rvert$ for any bijection $f: \mathbb R \to [0, 1]$. Similarly there are metrics on $(0, 1)$ and $[0, 1]$ that make them unbounded. But when we say $\mathbb R$ is unbounded, we're always going to be referring to the usual metric.

bitesizebo
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  • Thank you for this! Is the space $X$ a topological space, or do you mean an even more general notion of space? – kusa Apr 25 '19 at 13:49
  • A metric $d$ on $X$ will induce a topology on $X$. We call a subset $A \subset X$ open in a metric space if for all $x \in A$ there exists $\epsilon > 0$ such that ${y \in X : d(x, y) < \epsilon} \subset A$. Then these open sets form a topology on $X$. So every metric space has an associated topology. Do note that different metrics can generate the same topology, for example the $\ell_p$ norms on $\mathbb R^d$ (in fact this is partly what motivated the idea of a topology). – bitesizebo Apr 25 '19 at 17:59
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Cardinality is not defined in a way that requires it to preserve structure. You could have equally asked why $\Bbb Q$ is equinumerous with $\Bbb N$ despite being densely ordered, whereas $\Bbb N$ is not densely ordered.

The point is that given a set, there are many different structures we can place on it. Even if it is finite, there are different ways to structure it (on a set of size $n$ there are many partial orders, only some a linear, etc.).

The fact that $\Bbb R$ "explodes towards infinity" is the same as the fact that $(0,1)$ "explodes towards $1$". And we know very well that it is sometimes useful to add $\pm\infty$ to $\Bbb R$ and be able to talk about functions which are continuous there, or has a limit at these non-real points. How is that different from talking about functions on $(0,1)$ and asking if they admit some sort of limit or value at $0$ and $1$?

It doesn't.

This is one of many steps towards understanding that in abstract mathematics it is useful to remember that sometimes your naive intuition breaks, and the only way to correct for this is to work slowly with the definitions, one step at a time, until you've developed a new intuition. It is also a useful step in understanding that:

  1. Having the same cardinality doesn't mean that you have the same structure.
  2. Having similar structure doesn't mean that you have additional similarities not required by the structure preserved (in this case, $(0,1)$ and $\Bbb R$ are isomorphic in several ways, but not in a way that preserves all the natural structures on these sets).
  3. Infinite sets can be weird.
Asaf Karagila
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  • Thank you for your answer. It turns out that while further pursuing the matter, I came across another answer of yours which truly settled the matter for me. My question, in retrospect, is no good at encapsulating what I was wondering. https://math.stackexchange.com/a/40318/641016 is the answer I’m referring to! – kusa Apr 25 '19 at 13:55