We define $\lfloor x\rfloor$ by

$$\lfloor x\rfloor \in \mathbb{Z} \land \lfloor x\rfloor \leq x \land( \forall z \in \mathbb{Z}, z\leq x \Rightarrow z\leq\lfloor x\rfloor)$$

Prove or Disprove the following statement:

$$\exists x \in \mathbb{R}, \forall \epsilon>0,\exists\delta>0,\forall w \in \mathbb{R},|x-w|<\delta \Rightarrow|\lfloor x\rfloor - \lfloor w\rfloor| < \epsilon $$

How should I know that if this statement is true or false before I start to prove? I draw a graph, it seems false to me. Because no matter what $x$ I pick, I will always find a $\delta$ so that $|x-w|$ will be in $\delta$ range, but this doesn't guarantee that $|\lfloor x\rfloor - \lfloor w\rfloor| < \epsilon $.

But I have trouble of finding such $x,\delta$ in general

  • $\begingroup$ How should I properly say that $w$ is just barely less than $x$ ? Another thing is that when I prove the negation, $\exists x$ becomes $\forall x$, therefore, I have no right to pick $x$ anymore $\endgroup$
    – ElleryL
    Oct 31 '14 at 15:15
  • $\begingroup$ I don't understand your notation. What does "$>$" mean where you use it? $\endgroup$
    – MPW
    Oct 31 '14 at 15:16
  • $\begingroup$ @MPW, sorry, just fix it, type error $\endgroup$
    – ElleryL
    Oct 31 '14 at 15:17
  • $\begingroup$ There's another one close to the end. I guess that's a typo as well. $\endgroup$
    – MPW
    Oct 31 '14 at 15:18
  • 1
    $\begingroup$ The statement that you are trying to prove is that there is some $x$ such that the floor function is continuous at that point. Since the point is to show that some such $x$ exists, finding an explicit $x$ such as Mirko has done above is a straightforward approach. $\endgroup$
    – copper.hat
    Oct 31 '14 at 15:30

As MPW pointed out (and as is usual, and as more comments agree on that), $\lfloor x \rfloor$ is the usual floor function, largest integer not exceeding $x$. (That is exactly what that formula on top says: $\lfloor x \rfloor$ is an integer, it does not exceed $x$ and there is no larger $z$ with these properties).

So, given that, take $x=1/2$ and $\delta=1/4$ (no matter what is $\epsilon>0$. Then $w$ must be in the interval $(1/4,3/4)$ so $\lfloor w \rfloor = \lfloor x \rfloor = 0$, so $| \lfloor w \rfloor - \lfloor x \rfloor | = 0 < \epsilon$.

Which, just in case it is not clear, is a proof of your statament $$\exists x \in \mathbb{R}, \forall \epsilon>0,\exists\delta>0,\forall w \in \mathbb{R},|x-w|<\delta \Rightarrow|\lfloor x\rfloor - \lfloor w\rfloor| < \epsilon $$

  • $\begingroup$ so basically, when it says $\exists \delta, \forall w$ it means I choose a $\delta$, and all $w$ must be in this $\delta$ range, I was confused at beginning and thought that $w$ could be arbitrarily as far from $x$ as possible. Thanks so much for the help $\endgroup$
    – ElleryL
    Oct 31 '14 at 15:44
  • $\begingroup$ Well, yes $\forall w$ does indeed mean for all, but then the rest if it should be read: IF $w$ is $\delta$-close to $x$ THEN $| \lfloor w \rfloor - \lfloor x \rfloor | < \epsilon$. So, you do not need to worry what happens with $| \lfloor w \rfloor - \lfloor x \rfloor |$ in the case when $w$ is NOT $\delta$-close to $x$. $\endgroup$
    – Mirko
    Oct 31 '14 at 15:47

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