Find functions in the specific range that is one on one correspondence

I am confused about how to represent a function to fix one on one correspondence, and how to represent a function only correspondence but not onto. Here are the example questions, could someone give me help or a tip?

Find a function f : [1, 2) → (1, 2) that is 1-1 correspondence.

f: (0.5, 1) → (0, 1) that is 1-1 but not onto;

f: (0.5, 1) → (0, 1) that is 1-1 correspondence;

• Welcome to MSE. Please, show us your effort. Sep 20, 2017 at 18:03
• I do not understand how to represent a function to have 1-1 correspondence in this range. I am new in the mathmatic set. Sep 20, 2017 at 18:14
• I believe that You coud be a little confused with the first example. But what with other examples? Sep 20, 2017 at 18:22
• Sorry for reply late, I am in the class. It's the same things I am confused, because I don't understand "how" should I figure out the answer, and how should I give a function that fits the problem. Sep 20, 2017 at 20:24

For the first problem map 1 to 1.1, map 1.1 to 1.11, map 1.11 to 1.111
and so forth ad infinitum. For all the other points, map each of them
to themselves.

The other two problems are easy.

Find a function f : [1, 2) → (1, 2) that is 1-1 correspondence.

An answer is already provided by William Elliot.

f: (0.5, 1) → (0, 1) that is 1-1 but not onto;

Put $f(x)=x$ for each $x$ from $(0.5,1)$.

f: (0.5, 1) → (0, 1) that is 1-1 correspondence;

Put $f(x)=2x-1$ for each $x$ from $(0.5,1)$.

• Can you elaborate a little? I can kind of understand #3 there, but I'm new to a lot of the symbols used and the theory behind them. $f(x) = 2x-1$, plug in $0.5, 1$ and you end up with $0,1$; is it such that for $f : A -> B$, you're plugging in values from A to find B? Sep 27, 2017 at 1:28
• @gator Let $x$ is from $(0.5, 1)$, that is $0.5<x<1$. Then $0<f(x)=2x-1<1$, so $f(x)$ is from $(0,1)$. Next, if $x$ and $y$ both are from $(0.5, 1)$ and $x\ne y$ then $f(x)=2x-1\ne 2y-1=f(y)$, so $f$ is 1-1-map. At last, for each $t$ from $(0,1)$ there exists $x$ from $(0.5,1)$ such that $f(x)=t$ (namely, $x=\frac{t+1}2$), so $f$ is 1-1 correspondense. Sep 27, 2017 at 1:35
• Thank you. $f(x) = x$ is injective but not bijective because $0.5 < f(x) < 1$ and not $0 < f(x) < 1$ (there exists no $x$ in $f(x) = x$ between $0..0.5$. $f(x) = 2x - 1$ does have $x$ that yields $0..1$ thus it is bijective. Sep 27, 2017 at 1:53
• @gator Yes..... Sep 27, 2017 at 1:54
• Thanks so much, I was so confused about how to represent a "function" in the question, until my prof told me you can also use words to describe it. Btw, the answer is very detailed! Sep 27, 2017 at 15:38

Your question needs some clarification: do you need to find an $f$ that is 1-1, i.e. injective, or do you need to find a 1-1 correspondence between the intervals, i.e. a bijection?

The problem is pretty easy if it's the first, so let's assume the second.

The obvious difficulty is the $1$ in the interval $[1,2)$, so think about where it will map to. Since we must have $1 < f(1) < 2$, $f(1)$ partitions the range into two open intervals $(1, f(1))\cup(f(1),2)$. Now the problem is to find a bijection between $(1,2)$ and those two open intervals.

• Could you please give me a function(second type) that represent my questions? I feel I misunderstand how should I answer it. Sep 20, 2017 at 20:48