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By definition, I know I have to show there exists a $1$-$1$ onto map $f:A\to B$ I am pretty stuck on how to go to the process of proving this.

I understand the basic definitions of 1-1 and onto. I recall basic definitions of $1$-$1$ and onto from high school, I know a function is $1$-$1$ if it passes the horizontal and vertical line test and a function is onto if for all $Y$ there exists an $x$ in $X$ such that $f(x) = y$, but I am really not sure how to go about proving this at all. Could someone maybe point me in the direction of an example to help me understand each step in the process?

Martin Sleziak
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user604296
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1 Answers1

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Guide:

  • Construct a linear mapping from $[0,0.5)$ to $[0,2),$ call it $f_1$.

  • Construct a linear mapping from $[0.5, 1)$ to $[3,4]$, call it $f_2$.

  • Check that $f:[0,1) \to [0,2) \cup [3,4)$, $f(x) = \begin{cases} f_1(x) &, x< 0.5 \\ f_2(x)& ,x \ge 0.5\end{cases}$ is a bijection.

Siong Thye Goh
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  • I guess I don't understand your terminology. What is a linear mapping and how did you pick those ranges? Did you just split the interval [0,1) in half? – user604296 Oct 17 '18 at 04:34
  • I just choose to split it in half, you can split it in another way and it still works. Can you construct a line connecting $(0,0)$ (this is a coordinate) and $(0.5,2)$ (this is another coordinate). – Siong Thye Goh Oct 17 '18 at 04:36
  • Thanks for your time but I am just beyond lost with this material. Is there anywhere online you would recommend to learn how to construct a proof like this? – user604296 Oct 17 '18 at 04:49
  • I don't really have any idea about resources to find examples. I thought those links in the comments should help. Perhaps a textbook? what is the difficulty currently? constructing $f_i$? – Siong Thye Goh Oct 17 '18 at 05:02
  • We aren't using a textbook for this class. He just gave us this problem saying there will be a similar question on the test. I usually learn best from examples and deriving what to do from there but I cant find a full example online. – user604296 Oct 17 '18 at 05:07
  • To the proposer: The idea is "cutting & pasting": If $a<b$ then $[a,b)=P\cup Q$ where $P=[a,(a+b)/2)$ and $Q=[(a+b)/2,b).$ Note that $P,Q$ are disjoint. If $c<d\leq c'<d' $ there is a linear function $f_1$ that maps $P$ bijectively onto $[c,d)$ and a linear function $f_2$ that maps $Q$ bijectively onto $[c',d'). $ .....Any non-constant real linear function is 1-to-1 and maps half-open intervals onto half-open intervals... E.g. the graph of $f_1$ is the line in $\Bbb R^2$ thru the points $<a,c>$ and $<(a+b)/2,d>$.... Using $<$ and $>$ instead of $($ and $)$ for ordered pairs. – DanielWainfleet Oct 17 '18 at 07:17