3 ms·
> which is indeed false, you can also interpret the expression as: > lim[eps->0] ((1-eps) < 1) Assuming you don't mean some special notion of limit, I would
by gspr 7y ago
> which is indeed false, you can also interpret the expression as:
> lim[eps->0] ((1-eps) < 1)
Assuming you don't mean some special notion of limit, I would guess that by `(1-eps) < 1` you mean the function from the reals to `Y = {false, true}` that is defined as sending `x` to `true` if `1-x` is strictly less than `1`, and `false` otherwise. Let's call this function `f`. I assume you're endowing `Y` with the discrete metric?
If so, `f` does indeed have a limit from above that is `false` and a limit from below that is `true`. Where do you wanna go from here?
Edit: Corrected stupid wrong assertion about limit from below, d'oh.
- amelius 7y agoHave a look here: https://www.wolframalpha.com/input/?i=Limit%5BSign%5B1-x-1%5D%2C+x-%3E0%5D https://www.wolframalpha.com/input/?i=Limit%5BSign%5B1-x-1%5... This shows that the limit does exist from both sides (but is different from both sides).
- gspr 7y agoOops, sorry, that was a stupid blunder. Thanks for the correction.
- amelius 7y ago> If so, `f` does indeed have a limit from above that is `false` and a limit from below that is `true`. Where do you wanna go from here? That's the other way around. From above you get true, and from below you get false. Note that 0.999... represents the limit from above for 1-eps. Hence the result is true.
- gspr 7y ago> That's the other way around. From above you get true, and from below you get false. Yeah, my bad. > Note that 0.999... represents the limit from above for 1-eps. Hence the result is true. I mean you can define 0.999… like that if you want. You'll get 1. So what? Your detour via `f` provided nothing.