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a) Show that any open segment $(a,b)$ with $a<b$ has the same cardinality as $\mathbb{R}$.

b) Show that any closed segment $[a,b]$ with $a<b$ has the same cardinality as $\mathbb{R}$.


Thoughts:

Since $a<b$, $a,b$ are two distinct real number on $\mathbb{R}$, we need to show it is 1 to 1 bijection functions which map between $(a,b)$ and $\mathbb{R}$, $[a,b]$ and $\mathbb{R}$.

But we know $\mathbb{R}$ is uncountable, so we show the same for $(a,b)$ and $[a,b]$?

and how can I make use of the Cantor-Schroder-Bernstein Theorem? The one with $|A|\le|B|$ and $|B|\le|A|$, then $|A|=|B|$?

thanks!!

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Let's do a fundamentally different approach. Consider projecting from the open bottom half of the circle $x^2 + (y-1)^2 = 1$, as I drafted below.

enter image description here

In this, by projecting from $(0,1)$, we get a 1-1 correspondence from the angle $(0, -\pi) \to \mathbb{R}$. There is a clear 1-1 correspondence from $(0,1)$ and $(0, -\pi)$.

Thus we have our bijection from $(0,1)$ to $\mathbb{R}$. For $[0,1]$, we could use big theorems, or we could just modify this one. So let's "make space" by saying that whatever was associated to $0$ is now associated to $2$, what was associated to $1$ is now associated to $3$, $2$ now goes to $4$, and so on, effectively freeing up the real numbers $0,1$ by displacing the natural numbers. Send $0$ to $0$, and $1$ to $1$, and now we have a bijection $[0,1] \to \mathbb{R}$.

Nice, direct, and constructive.

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Hint: $(a,b)\subseteq[a,b]\subseteq\Bbb R$ so you only need to show an injection from $\Bbb R$ into $(a,b)$. Then the Cantor-Bernstein theorem can work its magic.

Show that every two open intervals have the same cardinality (exhibit a bijection between them), and so it is enough to show that there is an injection into $(0,1)$. Consider something along the lines of: $$x\mapsto\frac1{2+e^x}$$

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This can be done by a straightforward application of the Cantor-Schroeder-Bernstein theorem. All you have to do is find an injective function $f:(a,b)\to \mathbb [a,b]$ and then an injective function $g:[a,b]\to (a,b)$. Now, there is one screamingly obvious candidate for a injective function $f:(a,b)\to [a,b]$. As for the other direction, can you think of a function that will shrink the interval $[a,b]$ just a little bit so it fits into $(a,b)$? Think of the case $[-1,1]$ and $(-1,1)$ for intuition if needed.

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  • 1
    $\begingroup$ I just noticed that all he answers actually answer something entirely different than the question $\endgroup$ – Asaf Karagila Feb 12 '13 at 3:16
  • $\begingroup$ Perhaps details left to the reader? $\endgroup$ – leo Feb 12 '13 at 3:31
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As mentioned in Ittay Weiss's answer, we just need to find a bijection $\beta: [-1, 1] \to (-1, 1)$. Try this approach: we need to "push" the endpoint $-1$ back into the open interval $(-1, 1)$. Let's say it ends up at some $a_1 \in (-1, 1)$. Then now $a_1$ got squeezed out so let's move it to some $a_2 \in (-1, 1)$. Some $a_3 \in (-1, 1)$ then needs to make way for $\beta(a_2)$. Continuing in this fashion of displacing and replacing, we can "push" the endpoints of $[-1, 1]$ inside $(-1, 1)$ without touching most of the points.

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Consider the function $f:(0,1)\to \mathbb{R}$ defined as,

$$f(x)=\frac{1}{x}+\frac{1}{1-x}$$

Prove that $f$ is a bijective function.

Now, by previous posts, $(0,1)$ and $[0,1]$ have the same cardinality.

Consider the function $g:[0,1]\to[a,b]$, defined as,

$$g(x)=({b-a})x+a$$

Prove that $g$ is bijective function to conclude that $[0,1]$ and $[a,b]$ have the same cardinality as $\mathbb{R}$.

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