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Question:

Show that all the intervals in $\mathbb{R}$ are uncountable.

I have already proven that $\mathbb{R}$ is uncountable by using the following:

Suppose $\mathbb{R}$ is countable. Then every infinite subset of $\mathbb{R}$ is also countable. This contradicts the fact that $\mathbb{I} \subset \mathbb{R}$ is uncountable. Consequently, $\mathbb{R}$ must be uncountable.

However, how can I show that ALL the intervals in $\mathbb{R}$ are uncountable?

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  • $\begingroup$ Try to create a bijection between an interval and $\mathbb R$. $\endgroup$ – kennytm Mar 31 '15 at 22:40
  • $\begingroup$ Nitpick: you mean all non-trivial intervals. $\endgroup$ – GFauxPas Mar 31 '15 at 22:42
  • $\begingroup$ @Aaron: This is a perfectly reasonable question in an elementary set theory course. To be fair, it's less about real analysis than it is about (elementary) set theory. $\endgroup$ – Asaf Karagila Mar 31 '15 at 23:03
  • $\begingroup$ @AsafKaragila It seems all real analysis to me. But you're the man!! So that's fine! $\endgroup$ – Aaron Maroja Mar 31 '15 at 23:04
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Let $f_s:\mathbb{R} \to \mathbb{R}$ be a scaling function $x \mapsto sx$ for some positive scaling factor $s$; then $f_s$ maps the unit interval to an interval of length $s$. Then let $g_t:\mathbb{R} \to \mathbb{R}$ be a shifting function $x \mapsto x+t$ for some shift $t$. Then to show that an arbitrary interval from $a$ to $b$, $a \neq b$, is uncountable, note that $g_a \circ f_{b-a}$ maps the unit interval to the interval from $a$ to $b$. $f_s$ and $g_t$ are both bijective, so the interval from $a$ to $b$ must have the same cardinality as the unit interval. Therefore the interval from $a$ to $b$ is uncountable. Since $a$ and $b$ are arbitrary, every interval of $\mathbb{R}$ is uncountable. Q.E.D.

Note: I'm assuming a priori that we know $\mathbb I$ is uncountable, but since you used that in your prior proof, that seems OK.

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  • $\begingroup$ Are you not assuming that $(0,1)$ (the unit interval) is uncountable? :) $\endgroup$ – user860374 Mar 31 '15 at 22:58
  • $\begingroup$ Already added a note about that, but yes, I am assuming that. I figure it's OK because you used that to prove that $\mathbb{R}$ is uncountable in the first place. $\endgroup$ – calavicci - GoFundMonica Mar 31 '15 at 23:02
  • $\begingroup$ You added a note saying you are assuming $\mathbb{I}$ is uncountable, not $(0,1)$. I really do like this proof though, so I will definitely make use of this :). It's elegant and well written :). So what I will do, is prove that $(0,1)$ is uncountable (which I have done already on a separate occasion) and simply use that result within this proof as motivation :) $\endgroup$ – user860374 Mar 31 '15 at 23:06
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    $\begingroup$ Oh, OK, I think I see - you're asking about the distinction between $[0,1]$ and $(0,1)$; is this correct? If so, just note that the cardinality of $(0,1)$ is the cardinality of $[0,1]$ minus 2; since the cardinality of $[0,1]$ is an uncountable infinity, subtracting 2 still gives you an uncountable infinity, so $(0,1)$ is uncountable. $\endgroup$ – calavicci - GoFundMonica Mar 31 '15 at 23:11
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HINT: Find an injection from $\Bbb R$ into the interval $(0,1)$. Then show there is a bijection between any two intervals $(a,b)$ and $(c,d)$.

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Hint: Consider $\phi : \mathbb R \to (-1,1)$ defined by $$\phi(x) = \frac{x}{1 +|x|}$$ and notice that $f: (-1,1) \to (a,b)$ defined by $$f(x) = \frac{1}{2}\bigg((b-a)x + a+ b\bigg)$$ is a bijection.

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I'll show you geometrically a one-to-one correspondence between $(0,1)$ and $(0,2)$, showing that they have the same cardinality.

(0,1) to (0,2)

To figure out where $x$ goes to, draw a line from the point $P$ on top, through $x$ on the copy of $(0,1)$, until it hits the copy of $(0,2)$. Like this:

enter image description here

You see, we know how to pair up any two points because one line joins exactly one point of $(0,1)$ with one point of $(0,2)$.

In fact, you could do this to pair up any sets $(a,b)$ and $(c,d)$, or $[a,b]$ and $[c,d]$.

Now, to show that these are all uncountable, it suffices to provide a one-to-one correspondence between $(0,1)$ and $\mathbb R$.

So, what I'm going to do is take the interval $(0,1)$ and curve it up into a half-circle:

enter image description here

We do the same thing we did before. We have a point at the top; each point $x$ of $(0,1)$ is paired with the point $f(x)$ of $\mathbb R$ that is on the line joining the point at the top with $x$:

enter image description here

As we get closer and closer to the boundary of $(0,1)$, it gets stretched out more and more. It would be getting stretched out infinitely far at $0$ or $1$, but it's OK because those points aren't there! And we get to take this finite thing and distort it, a lot, to cover this infinite thing! (Since we're doing $(0,1)$, $0$ and $1$ aren't included in the set we're trying to pair up.)

Script and images stolen from here: http://www.artofproblemsolving.com/school/mathjams-transcripts?id=196

Now, one thing I haven't yet shown you is a one-to-one correspondence between $(0,1)$ and $(0,1]$ (or, for that matter, between either of those and $[0,1]$.) I haven't been able to find an image for this — I might edit this later if I find one. (Or I could draw it and photograph it with my phone.) However, hopefully I've convinced you that these are all uncountable, and that's all you really wanted to do, so this detail isn't essential.

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