This is kind of a more conceptual and perhaps silly question

But if I have an interval [0, 1)

I think of it as [0, 1] with 1 real removed from it

Is it nonsense to say there is a real X that is lesser than 1, but greater than all other reals lesser than 1

Because that is the interval [0, X]

Which would also be [0, 1)

And I’d ask what the midpoints are of both intervals

And then I repeat the procedure again

perhaps branching the question to snipping off a real from either end

(it’s also easy to show you can’t have [0,1) = [0,X] because the topology is wrong. [0,X] is closed, so it contains every limit point, but 1 is a limit point of [0,1) which it doesn’t contain)

@slack1 it’s correct to say that [0,1) has one fewer real than [0,1] (in a certain sense)

but as brendan says, there is no real that bounds [0,1) above smaller than 1

I see, so you’re saying my X is contradictory

yes

to see why, assume there is such an X

by assumption, X < 1

then 1 - X > 0

so then 1 > (1-X)/2 + X > X

but then (1-X)/2 + X must be in the interval, since it’s less than 1

but it’s bigger than X, contradicting our assumption that X was the biggest such number

ahh

the untamable reals

Analysis is gross and I don’t like it

Which probably why I explained it topologically instead of with inequalities lol

right, @brendan’s proof is much shorter and more elegant but requires topology instead of high school math