Help with using the Schroeder-Bernstein Theorem? Corresponding Counts [3 points] Prove that <m

Help with using the Schroeder-Bernstein Theorem?
Corresponding Counts [3 points]
Prove that $|\left\{x\in \mathbb{R}|0\le x\le 1\right\}|=|\left\{x\in \mathbb{R}|4.
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Terrance Phillips
Step 1
The injection $x\to x+5$ shows that we have an injection from [0, 1] into (4, 7), so by definition $|\left[0,1\right]\le |\left(4,7\right)|$.
Step 2
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 $\left(\frac{4}{7},1\right)\subseteq \left[0,1\right]$, so $|\left(4,7\right)|\le |\left[0,1\right]|$.
Now Cantor-Bernstein does the rest.
It's a handy tool to not have to give an exact bijection between these two sets.
Not exactly what you’re looking for?
Andy Erickson
Step 1
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:
$\left[0,1\right]↦\left(4,7\right)$, let's find a function that will map every input from [0, 1] to (4, 7). A function that would accomplish this would be $f\left(x\right)=x+5$. To show this you could make a simple diagram like: $0↦5\phantom{\rule{0ex}{0ex}}.5↦5.5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⋮\phantom{\rule{0ex}{0ex}}1↦6$
Step 2
For $\left(4,7\right)↦\left[0,1\right]$, a potential function could be: $f\left(x\right)=\frac{x}{4}-1$. Again, drawing a diagram like:
$5↦.25\phantom{\rule{0ex}{0ex}}⋮$
By showing $\left[0,1\right]↦\left(4,7\right)$ is one-to-one, then $|\left[0,1\right]|\le |\left(4,7\right)|$ and have shown $\left(4,7\right)↦\left[0,1\right]$ meaning $|\left(4,7\right)|\le |\left[0,1\right]|$ is also one-to-one, then we can conclude using Schröder-Bernstein that $|\left[0,1\right]|=|\left(4,7\right)|$.