To construct a bijection between the intervals $[a, b]$ (a closed interval) and $(C, d)$ (an open interval), we use a shifting and stretching method. Here's how we can construct such a bijection step by step:

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1. Shifting and Stretching

Mapping $[a, b] \to (C, d)$

We aim to find a function $f : [a, b] \to (C, d)$ such that:

• $f(a) = C$ (maps the left endpoint of $[a, b]$ to $C$),
• $f(b) = d$ (maps the right endpoint of $[a, b]$ to $d$),
• $f$ is bijective (one-to-one and onto).

The formula for such a function can be: $f(x) = C + (d - C) \frac{x - a}{b - a}.$

Explanation of the Formula:

• $(x - a) / (b - a)$ scales $x$ in $[a, b]$ to $[0, 1]$,
• Multiplying by $(d - C)$ stretches $[0, 1]$ to $[0, d - C]$,
• Adding $C$ shifts the result to the interval $(C, d)$.

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2. Inverse Function

Since $f$ is a bijection, we can find its inverse $f^{-1}: (C, d) \to [a, b]$: $f^{-1}(y) = a + (b - a) \frac{y - C}{d - C}.$

Explanation of the Inverse Formula:

• $(y - C) / (d - C)$ scales $y$ in $(C, d)$ to $[0, 1]$,
• Multiplying by $(b - a)$ stretches $[0, 1]$ to $[0, b - a]$,
• Adding $a$ shifts the result to the interval $[a, b]$.

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3. Composition of Functions

Let us confirm that the composition of $f$ and $f^{-1}$ is the identity function.

$f(f^{-1}(y)) = y$:

Substitute $f^{-1}(y)$ into $f(x)$: $f(f^{-1}(y)) = f\left(a + (b - a) \frac{y - C}{d - C}\right).$ Using the definition of $f(x)$: $f(f^{-1}(y)) = C + (d - C) \frac{\left(a + (b - a) \frac{y - C}{d - C}\right) - a}{b - a}.$ Simplify: $f(f^{-1}(y)) = C + (d - C) \frac{(b - a) \frac{y - C}{d - C}}{b - a}.$ $f(f^{-1}(y)) = C + (d - C) \frac{y - C}{d - C}.$ $f(f^{-1}(y)) = C + (y - C) = y.$

$f^{-1}(f(x)) = x$:

Substitute $f(x)$ into $f^{-1}(y)$: $f^{-1}(f(x)) = f^{-1}\left(C + (d - C) \frac{x - a}{b - a}\right).$ Using the definition of $f^{-1}(y)$: $f^{-1}(f(x)) = a + (b - a) \frac{\left(C + (d - C) \frac{x - a}{b - a}\right) - C}{d - C}.$ Simplify: $f^{-1}(f(x)) = a + (b - a) \frac{(d - C) \frac{x - a}{b - a}}{d - C}.$ $f^{-1}(f(x)) = a + (b - a) \frac{x - a}{b - a}.$ $f^{-1}(f(x)) = a + (x - a) = x.$

Thus, $f$ and $f^{-1}$ are inverses, confirming the bijection.

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Summary

The bijection between $[a, b]$ and $(C, d)$ is:

• Forward mapping: $f(x) = C + (d - C) \frac{x - a}{b - a}$,
• Inverse mapping: $f^{-1}(y) = a + (b - a) \frac{y - C}{d - C}$,
• Both compositions $f(f^{-1}(y)) = y$ and $f^{-1}(f(x)) = x$ are identities.

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Questions for Further Exploration

1. How would you modify the formula if $(C, d)$ were a closed interval instead of open?
2. Can you derive a bijection if $[a, b]$ and $(C, d)$ are flipped, i.e., mapping $(C, d)$ to $[a, b]$?
3. How does the choice of intervals $[a, b]$ and $(C, d)$ affect the slope of the function $f(x)$?
4. What happens to the bijection if $C > d$, i.e., the interval is decreasing?
5. Can you apply the shifting and stretching method to higher-dimensional spaces, like rectangles or cubes?

Tip:

When verifying a bijection, always confirm both the forward and inverse compositions to ensure they yield the identity function.