Topology: Bijection with Open intervals

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Discussion Overview

The discussion centers around proving the equivalence of two open intervals (a,b) and (c,d) through the construction of a bijective function. Participants explore the properties of the function f(x) = ((d-c)/(b-a))*(x-a)+c, specifically its one-to-one and onto characteristics, and the implications of this proof in the context of topology.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks guidance on proving that the function f is a bijection between the intervals (a,b) and (c,d).
  • Another participant suggests that to prove f is one-to-one, one must show that if f(x) = f(y), then x must equal y, and provides algebraic reasoning for this.
  • To demonstrate that f is onto, it is proposed that one must show for any y in (c,d), there exists an x in (a,b) such that f(x) = y, and provides a method for solving this.
  • A later reply introduces the idea of modifying the bijection to show equivalence between (a,b) and (c,d] by using a countable subset, suggesting a more complex relationship between different types of intervals.

Areas of Agreement / Disagreement

Participants generally agree on the approach to proving the bijection, with some providing additional insights on extending the concept to other interval types. No significant disagreements are noted, but the discussion remains focused on the mathematical proof rather than reaching a consensus on broader implications.

Contextual Notes

Participants do not explicitly address potential limitations or assumptions in their arguments, nor do they clarify the implications of the bijection beyond the immediate proof.

MathColie
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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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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.
 
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:
OK Thanks so much. That was very helpful, I understand this now :smile:
 

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