How to prove two closed intervals are equinumerous?

Question

Prove that $[0,1] \approx [-\pi,e^2]$

(The notation $\approx$ is used to denote equinumerous)

I know to prove two closed intervals are equinumerous I need to show a bijective function that will map from one interval to the other but I cannot seem to find the correct mapping.

Can you find a linear function that maps $0$ to $-\pi$ and $1$ to $e^2$? — Ethan Bolker, Mar 29 '17 at 23:52
$f(x)=x(e^2+\pi )-\pi$ I have come up with this @EthanBolker — ECollins, Mar 29 '17 at 23:56
Hint: Think of the line joining $(0,-\pi)$ and $(1,e^2)$. Write the equation of this line in the form $y=mx+c$ — learning_math, Mar 29 '17 at 23:57

Yunus Syed · Answer 1 · 2017-03-30T03:43:01.783

0

Hint: $[-\pi, e^2] \cong [0, e^2 + \pi]$

Edit: I don't understand why I received a downvote so here's the full solution: The linear function from $[0,1] \to [0, e^2 + \pi]$ is just multiplication by $e^2 + \pi$ and is clearly a bijection. We then compose this with the map which subtracts $\pi$. This composition is a composition of bijections and is therefore a bijection.

edited Mar 30 '17 at 03:43

answered Mar 30 '17 at 00:01

Yunus Syed

1,737

2

I think your hint is to a more sophisticated/complex solution (which you only provided just now) than to the more straightforward "find a linear function". And that simpler solution was already hinted at. (Not my downvote). – Ethan Bolker Mar 30 '17 at 03:52

How to prove two closed intervals are equinumerous?

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