Topology: Bijection with Open intervals

In summary, the conversation discusses proving the equivalence of any two open intervals and how to show that a linear function is a bijection. The process of proving this is explained through the concepts of one-to-one and onto functions. The conversation also mentions the possibility of modifying the bijection to prove equivalence for other interval variations.
  • #1
MathColie
2
0
Good Morning,
I am trying to prove that any 2 open intervals (a,b) and (c,d) are equivalent.
Show that f(x) = ((d-c)/(b-a))*(x-a)+c is one-to-one and onto (c,d).

a,b,c,d belong to the set of Real numbers with a<b and c<d.
Let f: (a,b)->(c,d) be a linear function which i graphed to help me visualize, restricting the domain to (a,b). I know I have to prove that f is a bijection but I am not sure how to do this.
I know that I am supposed to show that any two open intervals are equivalent, even if the intervals are different lengths, so the open interval (3,4) is equivalent to the open interval (1,8), (3,4) is a subset of (1,9).

I would appreciate any help with this if you can get me on the right track...thanks so much!:confused:
 
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  • #2
MathColie said:
Good Morning,
I am trying to prove that any 2 open intervals (a,b) and (c,d) are equivalent.
Show that f(x) = ((d-c)/(b-a))*(x-a)+c is one-to-one and onto (c,d).

a,b,c,d belong to the set of Real numbers with a<b and c<d.
Let f: (a,b)->(c,d) be a linear function which i graphed to help me visualize, restricting the domain to (a,b). I know I have to prove that f is a bijection but I am not sure how to do this.
I know that I am supposed to show that any two open intervals are equivalent, even if the intervals are different lengths, so the open interval (3,4) is equivalent to the open interval (1,8), (3,4) is a subset of (1,9).

I would appreciate any help with this if you can get me on the right track...thanks so much!:confused:

You prove that it is a bijection by proving just what you said: that it is one-to-one and onto.

one-to-one: if f(x)= f(y) then x= y: if ((d-c)/(b-a))*(x-a)+c= ((d-c)/(b-a))*(y-a)+c just do the algebra1

onto: if y is in (c,d), you must show that there exist an x in (a, b) so that f(x)= y. That is, if y= ((d-c)/(b-a))*(x-a)+c, with c< y< d, solve for x and show that a< x< b.
 
  • #3
Once you've done that, then for fun you can modify that bijection to prove that (a,b) and (c,d] (or variations of that) are equivalent. Just take an infinite countable subset of each (e.g. a rational subset) so that d maps into the subset of (a,b) and then all the elements of the subset get "shifted over" by index, and all the other elements mapped using your bijection.
 
Last edited:
  • #4
OK Thanks so much. That was very helpful, I understand this now :smile:
 

1. What is topology?

Topology is a branch of mathematics that studies the properties and relationships of geometric shapes and spaces, such as the properties of continuity, connectedness, and compactness.

2. What is a bijection?

A bijection is a function between two sets where each element in one set is paired with exactly one element in the other set, and vice versa.

3. How is topology related to bijections with open intervals?

In topology, bijections with open intervals refer to a specific type of mapping between two topological spaces where open intervals in one space are paired with open intervals in the other space. This type of mapping is useful in understanding the topological properties and relationships between these spaces.

4. What are open intervals?

Open intervals are sets of real numbers that do not include their endpoints. For example, (0, 1) is an open interval that includes all real numbers between 0 and 1, but not 0 or 1 themselves.

5. How are bijections with open intervals used in topology?

Bijections with open intervals are used in topology to establish a one-to-one correspondence between points in two topological spaces. This allows for a deeper understanding of the topological properties and relationships between these spaces, such as continuity, connectedness, and compactness.

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