1

I am studying Discrete math at the moment and having trouble finding bijections and building functions.. two problems for example: $$[2, 5) \to (4, \infty)$$ and another example: $$[0,1] \to (0,1)$$

Is there a good strategy?
thanx...

N. F. Taussig
  • 76,571
Itamar Silverstein
  • 67
  • This is subtle. Possible duplicate of http://math.stackexchange.com/questions/28568/bijection-between-an-open-and-a-closed-interval – Ethan Bolker Dec 31 '15 at 15:49

1 Answers1

2

The idea: to get rid of the problematic extra (border) points take countable subsets of the unproblematic open intervals and use the trick of the Hilbert's Hotel. Example:

$$f:[0,1)\longrightarrow(0,1)$$ $f(x) = x$ except for: $$f(0)=1/2,$$ $$f(1/n)=1/(n+1),$$

Martín-Blas Pérez Pinilla
  • 41,546
  • 4
  • 46
  • 89
  • [2,5)→(4,∞) how about this problem? the strategy does not seem to work for harder problems... im lost :( – Itamar Silverstein Dec 31 '15 at 16:27
  • @ItamarSilverstein, can you find a bijection between $(2,5)$ and $(4,\infty)$? Between $(0,1)$ and $(0,\infty)$? – Martín-Blas Pérez Pinilla Dec 31 '15 at 16:30
  • @ItamarSilverstein, example: $x\to 1/x$ is a bijection between $(0,1)$ and $(1,\infty)$. – Martín-Blas Pérez Pinilla Dec 31 '15 at 16:32
  • so how do you find [2,5)→(4,∞) ? I still can't see that.... not everyone is a genius,,, – Itamar Silverstein Dec 31 '15 at 16:54
  • @ItamarSilverstein, $x\mapsto 3+1/x$ is a bijection between $(0,1)$ and $(4,\infty)$. Can you continue? – Martín-Blas Pérez Pinilla Dec 31 '15 at 16:57
  • @ItamarSilverstein, do a chain of bijections between open intervals. – Martín-Blas Pérez Pinilla Dec 31 '15 at 17:00
  • but math is about finding algorithms. can you give me strong alogrithms for find that? for example. in this particular example you examine the function 1/x and get that 1/x + 3 is mapping (0,1) to (4, ∞), but what is the next step? you need to find a bijection [2, 5) to (0, 1), what is the intuition? – Itamar Silverstein Dec 31 '15 at 18:08
  • 1
    Math is about solving problems, and for some of them there is no algorithm. However, we have the ability to recognize patterns and develop intuitions and successful strategies. Any bounded interval $(a,b)$ is can be bijected with another bounded interval $(c,d)$ by a linear function. Knowing a bijection $(0,1) \to (1,+\infty)$ gives you bijections between $(0,1)$ and $(-\infty, a)$ and $(b,+\infty)$, any $a,b$. So now you can show that there's a bijection between any two open intervals, bounded or unbounded. It remains to know bijections between $(0,1), (0,1], [0,1), [0,1]$. – BrianO Dec 31 '15 at 19:24