One easy question about limit of function at a point. In the book what I read says: Let $X\subset \mathbb{R}, $ $f: X\rightarrow \mathbb{R}$, $E\subset X$, $x_0$ be an adherent point of $E$ and $L\in \mathbb{R}$, we say that $f$ converges to $L$ at $x_0$ in $E$ iff for every $\varepsilon >0$ there exists a $\delta>0$ such that $|f(x)-L|\le \varepsilon$ when $f$ is restricted to all the $x\in E$ such that $|x-x_0|<\delta$ and we write it $\lim_{x\rightarrow x_0;x\in E}f(x)=L$. Which is equivalent to the following statement for all the sequences $(a_n)$ which converges to $x_0$ and $a_n \in E$ for all $n$, so $(f(a_n))$ converges to $L$.

So my question is regard to this equivalent definition, for example if I'd like to show that the limit laws hold the result becomes trivial when change everything a sequences (I've already proven all the limit laws for sequences), Am I right?

For example, assuming that either $f,g$ are function such that $\lim_{x\rightarrow x_0;x\in E}f(x)=L$ and $\lim_{x\rightarrow x_0;x\in E}g(x)=M$ makes sense and I want to show that $\lim_{x\rightarrow x_0;x\in E}f(x)g(x)=LM$. So it's as simple as pick some arbitrary sequence $(a_n)$ which takes its element from $E$ and converges to $x_0$ that always is possible since $x_0$ is an adherent point. By the equivalence we already know that $f(a_n)\rightarrow L$ and $g(a_n)\rightarrow M$. Thus $fg(a_n)=f(a_n)g(a_n)\rightarrow LM$ and since $(a_n)$ was completely arbitrary then $\lim_{x\rightarrow x_0;x\in E}f(x)g(x)=LM$.

this is correct doesn't it?

Similarly if we have the function $\begin{cases} 1, & x\in \mathbb{Q} \\ 0, & x\notin \mathbb{Q}\end{cases}$

And I'd like to show that is not continuous everywhere it will sufficient to show that for all $x_0\in \mathbb{R}$ the limit is always undefined.

Suppose for the sake of contradiction that there exists a $x_0\in \mathbb{R}$ where the limit exist. So, there is a sequence $(q_n)$ such that $q_n \in \mathbb{Q}$ and $q_n\rightarrow x_0$ (this is always possible since each real number is an adherent point of the rational numbers), then $f(q_n)\rightarrow 1$. But we can also choose the sequence $(x_0+\sqrt{2}/n)$, which converges to $x_0$, and $f((x_0+\sqrt{2}/n)\rightarrow 0$. Since $0\not= 1$ this contradicts the uniqueness of the limit of any sequence. Hence its limit is always undefined and in particular is not continuous at any real number.

All this is correct doesn't? Because in a lot of books and notes I've seen more unwieldy arguments using $\varepsilon-\delta$.

Thanks in advance

  • $\begingroup$ At the end of the first paragraph, you probably mean $a_n\in E$... $\endgroup$ – zozoens Feb 3 '14 at 21:58
  • $\begingroup$ Yes that's right, Thanks $\endgroup$ – Jose Antonio Feb 3 '14 at 21:59
  • $\begingroup$ Otherwise, everything looks good to me, your reasoning is correct. $\endgroup$ – zozoens Feb 3 '14 at 22:00
  • $\begingroup$ Thanks, this equivalence definition IMO makes a lot of things much more easy to handle. I don't see why is the advantage of the large and cumbersome proof of the discontinuity of Dirichlet's function from the $\varepsilon-\delta$ point of view. $\endgroup$ – Jose Antonio Feb 3 '14 at 22:09

It looks good, for the most part. The only issue with your proof that the given piecewise function is nowhere continuous is that you still need to show that $x_0+\frac{\sqrt{2}}n$ is irrational for all sufficiently large $n.$ You could instead show/note/use the fact that every real number is an adherent point of the irrationals, so there is also a sequence of irrationals converging to $x_0,$ from which we can draw the conclusion you stated.

As an alternative approach, note that the given function's range is $\{0,1\}.$ This makes things simpler, as it turns out.

Claim: If $f:\Bbb R\to\{0,1\}$ is continuous at a point $x_0$, then it is constant in some open interval about $x_0$.

Proof: Suppose by way of contradiction that $f$ is not constant in any such open interval. In particular, then, for each $n\ge 1,$ there exists some $x_n\in\left(x_0-\frac1n,x_0+\frac1n\right)$ such that $f(x_n)\ne f(x_0).$ Take $\epsilon=\frac12.$ By construction, $x_n\to x_0,$ so $f(x_n)\to f(x_0)$ by continuity, so there is some $N$ such that $|f(x_n)-f(x_0)|<\epsilon$ for all $n>N.$ But since $f:\Bbb R\to\{0,1\},$ then for all $n\ge 1,$ we have $|f(x_n)-f(x_0)|=1,$ so $1<\epsilon=\frac12.$ Contradiction. $\Box$

We can easily generalize the above result to functions $f:\Bbb R\to F$ for any non-empty finite $F\subseteq\Bbb R.$ This is trivial to prove when $F$ consists of a single point, and otherwise, we take $$\epsilon=\frac12\min\{|y-z|:y,z\in F\text{ and }y\ne z\}$$ rather than $\epsilon=\frac12.$ Since $F$ is finite, then it isn't difficult to show that $\epsilon>0$--we need only show that $\{|y-z|:y,z\in F\text{ and }y\ne z\}$ is a finite set of positive numbers and we're basically done--and from there, the proof proceeds in the same sort of way.

  • $\begingroup$ Maybe the easiest way is: let $\varepsilon > 0$, then $\exists q\in \mathbb{Q}$ such that $\frac{x_0-\varepsilon}{\sqrt2}< q<\frac{x_0+\varepsilon}{\sqrt2}$. We claim that is always possible to choose a $q\not= 0$. Suppose $q=0$, so $0<\frac{x_0+\varepsilon}{\sqrt2}$ and exists a $q'$ such that $0<q'<\frac{x_0+\varepsilon}{\sqrt2}$ thus $q'\not=0$ and also works. Hence $x_0-\varepsilon< \sqrt{2}q< x_0+\varepsilon$ where $q\in \mathbb{Q}\backslash \{0\}$. Then $\sqrt2q$ is an irrational number since otherwise $1/q (\sqrt2q) = \sqrt2 \in \mathbb{Q}$. $\endgroup$ – Jose Antonio Feb 4 '14 at 1:09
  • $\begingroup$ This shows that the real numbers are adherent to the irrational. And we can construct a sequence such as $(i_n)\rightarrow x_0$ and $i_n\in \mathbb{R} \backslash \mathbb{Q}$ and... (thanks for the claim). $\endgroup$ – Jose Antonio Feb 4 '14 at 1:14
  • $\begingroup$ Yes, indeed! That shows just what you want. Nicely done. (You're welcome.) $\endgroup$ – Cameron Buie Feb 4 '14 at 1:52

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