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practice question from textbook:

All that my professor told us about this theorem was that if $|A| \le |B|$ and $|B| \le |A|$, then $|A| = |B|$.

I have no idea how to use this in this example from my book, how to use this theorem in general. Lectures have been no help.

Any kind of explanation on how to approach problems like these, how to use this theorem would be appreciated.

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Tuneum
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    Construct an injection from the closed interval $[0,1]$ to the open interval $(4,7)$, then construct an injection going the other way (i.e. going from the open interval $(4,7)$ to the closed interval $[0,1]$). For instance, consider linear injections (in fact, consider linear bijections). – SurfaceIntegral Oct 22 '21 at 00:53
  • Related https://math.stackexchange.com/questions/28568/bijection-between-an-open-and-a-closed-interval –  Oct 22 '21 at 01:08

2 Answers2

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The injection $x \to x+5$ shows that we have an injection from $[0,1]$ into $(4,7)$, so by definition $|[0,1] \le |(4,7)|$.

The injection $x \to \frac{x}{7}$ shows that we have an injection from $(4,7)$ into $[0,1]$ (the image of $(4,7)$ is $(\frac{4}{7},1)\subseteq [0,1]$), so $|(4,7)| \le |[0,1]|$.

Now Cantor-Bernstein does the rest.

It's a handy tool to not have to give an exact bijection between these two sets.

Henno Brandsma
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The main strategy for these kinds of problems is that you want to show that one set is one-to-one and the other is one-to-one as well. You show this by finding a function the maps from one set to the other.

For this question let's say that the interval $[0,1]$ is $A$, and the other $(4,7)$ is $B$. Showing that $|A| \le |B|$ is saying that this function is one-to-one. To show this:

$[0,1] \mapsto (4,7)$, let's find a function that will map every input from $[0,1]$ to $(4,7)$. A function that would accomplish this would be $f(x) = x + 5$. To show this you could make a simple diagram like: $$0\mapsto5 \\ .5\mapsto5.5\\ \\ \vdots \\1\mapsto6$$

For $(4,7) \mapsto[0,1]$, a potential function could be: $f(x) = \frac{x}{4}-1$. Again, drawing a diagram like: $$5 \mapsto .25 \\ \vdots$$

By showing $[0,1] \mapsto (4,7)$ is one-to-one, then $|[0,1]| \le |(4,7)|$ and have shown $(4,7) \mapsto[0,1]$ meaning $|(4,7)| \le |[0,1]|$ is also one-to-one, then we can conclude using Schröder-Bernstein that $|[0,1]|=|(4,7)|$.

James
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