how to prove [0,1] ~ [0,2]?

[raw latch]

how can i prove that [0,1] and [0,2] have the same cardinality?

I know that i need to define a bijection between them but like how?

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[hollow bear]

The bijection is really simple

What's the first function you can think of that has a domain of [0, 1] and a range of [0, 2]?

[raw latch]

oh, its just f(x)=2x

[hollow bear]

yep lol

you can imagine stretching out [0, 1] onto [0, 2]

[raw latch]

what do you mean?

[hollow bear]

imagine taking a line between 0 and 1 on the real number line, grabbing the right endpoint, and stretching it out until the right endpoint is at 2

[raw latch]

does that indicate they have the same cardinality?

[hollow bear]

Its just a geometrical interpretation of f(x) = 2 x

Scaling by 2

It does indicate they have the same cardinality, because that mapping is bijective (every point on [0, 2] is mapped to by exactly one point on [0, 1])

[lyric iron]

more generally you can show $[a,b]$ and $[c,d]$ are always equinumerous by constructing the line segment between them

[low helm]

Omegabet_

[raw latch]

can you give an example, i dont really understand

[lyric iron]

draw it

draw a line with domain [a,b] and image [c,d], then determine what the equation must be

it'll be some function $f(x)=mx+p$, which you can clearly show is a bijection

[low helm]

Omegabet_

[raw latch]

so for example these two intervals [0,4] and [6,8]. How can I determine the equation?

this is also why is was confused with [0,1] ~ [0,2] at first, it looks like they cant have the same cardinality

[lyric iron]

pick the line with endpoints (0,6) and (4,8)

that, by construction, is the diagonal of $[0,4]\times[6,8]$

[low helm]

Omegabet_

[lyric iron]

the line has slope $\frac{8-6}{4-0}=\frac{1}{2}$ and a y intercept of $6$, so $f: [0,4]\to [6,8]$ is given by $f(x)=\frac{1}{2}x+6$