Analysis, Proof about, f being continuous, bijective

In summary: Q->R is the same as g:Q->R, but with the addition that for every y in R there is a unique x in Q such that f(x) = g(y). This is important because if we could find a function h:Q->R such that f:Q->R is continuous and h:Q->R is also a bijection, then we would have a function which is both continuous and a bijection. Unfortunately, this is impossible, so we can safely say that f:Q->R is continuous and onto.In summary, if f is a non-decreasing function on the rationals and f restricted to Q is a bijection, then f:Q->R is
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Let f: R[tex]\rightarrow[/tex]R be a non-decreasing function. Suppose that f maps Q to Q and f: Q[tex]\rightarrow[/tex]Q bijection. Prove that f: R[tex]\rightarrow[/tex]R is continuous, one to one and onto.

Hello everyone, I have been staring at this statement for a while now and I just don't understand it, hence I can't even begin to prove it. Can someone explain to me in different words what am I being asked to do. Is f the same same function and Q[tex]\rightarrow[/tex]Q is somehow "inside" R[tex]\rightarrow[/tex]R. I don't understand how f could be the same if R and Q don't have the same cardinality, or in other words-I am lost. Any type of help is greatly appreciated.
 
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  • #2
f maps Q to Q means that if q is in Q, f(q) is also in Q.
I am assuming (unless otherwise noted) that Q stands for the rationals.
 
  • #3
Yes, Q=rationals, R=reals; but how and why is mapping of Q to Q relevant here, since I am later asked about map of function R to R?
 
  • #4
It is relevant because it is a critical hypothesis for the problem. You are also given that f is non decreasing and that f restricted to Q is a bijection. Think about what the map f restricted to Q looks like. Where does f(q) go to?

Now, where does f map the irrational points to?
 
  • #5
Since I=irrationals are uncountable and so are reals I guess we could create a bijection between the two (not sure if I am correct here), but I suppose I would be mapped to R.
 
  • #6
Ah, I deleted my previous post, what I said was incorrect!. But... let me try to be more helpful: try this outline of a proof (if you like):

1) show that the restriction of f to Q is continuous (treat Q as a metric space on its own and apply the epsilon delta definition).

2) show that there exists a unique extension of f:Q->Q to f:R->R such that f is continuous on R (proving this is fairly easy via contradiction).

3) show that f being non-decreasing on Q and continuous on Q implies it is this candidate function (step 2 isn't entirely necessary, but it should help provide a lot more intuition to make this last leap).

I'm quite drowsy, so if something sounds confusing tell me.
 
  • #7
I'm sorry, but I'm not familiar with the term "unique extension" what do you mean by that?
 
  • #8
Robert IL said:
I'm sorry, but I'm not familiar with the term "unique extension" what do you mean by that?

Oh no prob. An extension of a function f:Q->Q to a function g:R->R is g such that for every x in Q, f(x) = g(x).
 

1. What is continuity in relation to functions?

Continuity is a mathematical concept that describes the behavior of a function. A function is considered continuous if it has no sudden jumps or breaks in its graph, meaning that the input and output values change smoothly and without interruption.

2. How do you determine if a function is continuous?

To determine if a function is continuous, you must check three conditions: 1) the function is defined at the point in question, 2) the limit of the function at that point exists, and 3) the limit and the function value at that point are equal. If all three conditions are met, the function is considered continuous at that point.

3. What is bijectivity?

Bijectivity is a property of a function where every element in the range of the function corresponds to exactly one element in the domain. This means that every output has a unique input, and every input has a unique output. A bijective function is both injective (one-to-one) and surjective (onto).

4. How can you prove that a function is bijective?

To prove that a function is bijective, you must show that it is both injective and surjective. To show injectivity, you can use the horizontal line test, where no horizontal line intersects the graph of the function more than once. To show surjectivity, you can use the vertical line test, where every vertical line intersects the graph of the function at least once. If both tests pass, the function is bijective.

5. What is the significance of proving a function is continuous and bijective?

Proving a function is continuous and bijective is important because it guarantees that the function has a well-behaved inverse, meaning that you can easily find the input value from a given output value. This makes it easier to solve equations involving that function, and also allows for certain simplifications and generalizations in mathematical proofs and applications.

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