Here's the question you clicked on:
swissgirl
@KingGeorge
Its my weekly question :)
If you dont like this one I can give you another one
It's been a little while since I've done actual analysis :P
Ya so let me get you smth easier
Well, the first part is pretty easy. It's basically just the definition of "not uniformly continuous"
BTW Congratulations. Where is ur offer from?
U of Rochester. I'm still waiting to hear back from most schools I applied to. Unfortunately, I also just recently got a rejection from Princeton :(
awwwwww :( That is sad They are stupiidddddd
Ah. This new one I can do. It's just the IVP (intermediate value theorem). To be honest, I wasn't expecting them to accept me.
Let's see how much I remember. Last time proved this, we started with the assumption that \(f(a)f(b)\le 0\). In this particular case, if \(f(a)=0\) or \(f(b)=0\), we're done. If \(f(a)f(b)<0\), then one of \(f(a),f(b)\) is negative, and the other positive. At this point, we had another theorem that proved there was then some \(c\in[a,b]\) s.t. \(f(c)=0\). Then define \(g(x)=f(x)+y\). Then, from what we just proved, it immediately follows that \(g(c)=y\). If we started with \(g(b)\le y\le g(a)\), take \(f(x)=g(x)-y\), so \(f(b)\le 0 \le g(a)\). From this we get that \(f(b)f(a)\le 0\) and we're done.
However, I cannot immediately remember how we proved that if \(f(a)f(b)\le0\), then there was some \(c\in[a,b]\) s.t. \(f(c)=0\).
hmmm gonna check my notes
hmmm I found a proof in my book using the supremum
Well, I'm sorry, but I've got to go now. I need to get up early tomorrow, so I need to head to bed. Have a good night, and good luck with this!