1
$\begingroup$

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;

$\endgroup$
4
  • $\begingroup$ Welcome to MSE. Please, show us your effort. $\endgroup$ Commented Sep 20, 2017 at 18:03
  • $\begingroup$ I do not understand how to represent a function to have 1-1 correspondence in this range. I am new in the mathmatic set. $\endgroup$
    – Saikorin
    Commented Sep 20, 2017 at 18:14
  • $\begingroup$ I believe that You coud be a little confused with the first example. But what with other examples? $\endgroup$ Commented Sep 20, 2017 at 18:22
  • $\begingroup$ 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. $\endgroup$
    – Saikorin
    Commented Sep 20, 2017 at 20:24

3 Answers 3

2
$\begingroup$

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.

$\endgroup$
0
1
+50
$\begingroup$

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 $(0.5,1)$.

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – gator
    Commented Sep 27, 2017 at 1:28
  • $\begingroup$ @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. $\endgroup$ Commented Sep 27, 2017 at 1:35
  • $\begingroup$ 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. $\endgroup$
    – gator
    Commented Sep 27, 2017 at 1:53
  • $\begingroup$ @gator Yes..... $\endgroup$ Commented Sep 27, 2017 at 1:54
  • 2
    $\begingroup$ 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! $\endgroup$
    – Saikorin
    Commented Sep 27, 2017 at 15:38
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Could you please give me a function(second type) that represent my questions? I feel I misunderstand how should I answer it. $\endgroup$
    – Saikorin
    Commented Sep 20, 2017 at 20:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .