Proving that $f: E\to \mathbb R$, defined by $f(x)=\frac{1}{x-x_0}$ is not uniformly continuous on $E$.

Question

Let $E\subset \mathbb R$ and $x_0\notin E$ is a limit point of $E$.

It is to be proven that $f: E\to \mathbb R$, defined by $f(x)=\frac{1}{x-x_0}$ is not uniformly continuous on $E$.

I tried to prove it as follows:

We fix an $\epsilon \gt 0$. Since $x_0$ is a limit point of $E$, for any $\delta\gt 0, \exists y\in E$ such that $|y-x_0|\lt \delta/4$ and $ |y-x_o|\lt \frac{1}{\epsilon+\frac 4{\delta}+|\epsilon-\frac 4{\delta}|} $
and

$\exists x\in E$ such that $|x-x_0|\lt \frac{3\delta}4$ and $|x-x_0|\gt \frac{1}{\frac 12(\epsilon+\frac 4\delta)+\frac 12|\epsilon-\frac 4\delta|}\tag{1}$

It follows that $|f(y)|\gt \epsilon+\frac 4{\delta}+|\epsilon-\frac 4{\delta}|$ and $|f(x)|\lt \frac 12(\epsilon+\frac 4{\delta})+\frac 12|\epsilon-\frac 4{\delta}|$

It follows that: $|x-y|\le |x-x_0|+|y-y_0|\lt 3\delta/4+\delta/4=\delta$ and that $|f(y)|-|f(x)|\gt \frac 12(\epsilon+\frac 4{\delta})+\frac 12|\epsilon-\frac 4{\delta}|=\max\{\epsilon, \frac 4\delta\}\gt \epsilon\implies |f(y)-f(x)|\gt \epsilon$. Since $\delta\gt 0$ is arbitrary, it follows that $f$ is not uniformly continuous on $E$.

Is my proof correct? Thanks.

Edit: In $(1)$: $x$ has been chosen in such a way that it remains in $(x_0-3\delta/4, x_0+3\delta/4)$, considering that $\max\{\epsilon,\frac 4\delta\}=\frac 12 (\epsilon +\frac 4\delta)+\frac 12|\epsilon -\frac 4\delta|\gt \frac 4\delta\implies \frac 1{\max\{\epsilon,\frac 4\delta\}}\lt \frac \delta 4$.
Now there are two cases:
1) $|x-x_0|\le \frac {\delta }4$
In this case, we have: $\frac 1{\max\{\epsilon,\frac 4\delta\}}\lt |x-x_0|\le \frac{ \delta}4\lt \frac {3\delta }4$
2) $|x-x_0|\gt \frac{\delta }4$.
In this case, we have $\frac \delta 4\lt |x-x_0|$ and $\frac 1{\max\{\epsilon,\frac 4\delta\}}\lt |x-x_0|\lt \frac{3\delta }4$.
That is, in both the cases we have $\frac 1{\max\{\epsilon,\frac 4\delta\}}\lt |x-x_0|\lt \frac{3\delta }4$ and this validates our choice of $x$ in $(1)$.

Answer 1

Let's try to solve the more general problem.

Suppose that $E\subset \mathbb{R},\ x_0\notin E$ and $x_0$ is a limit point of $E$. Suppose we are given the function $f:E\to \mathbb{R}$ with the following property: in any deleted neighborhood $f$ is unbounded, i.e. for any $r>0$ the function $f(t)$ is unbounded on $E\cap ((x_0-r,x_0+r)\setminus \{x_0\})$.

One can check that your function $E\mapsto \mathbb{R}$ defined by $t\mapsto \dfrac{1}{t-x_0}$ has this property.

Take any $\delta>0$ consider the neighborhood $W:=E\cap ((x_0-\delta/2,x_0+\delta/2)\setminus \{x_0\})$. Pick a point $x_2\in W$. Since $f$ is unbounded on $W$ then one can find $x_1\in W$ such that $|f(x_1)|>1+|f(x_2)|$.

Therefore, for $\forall \delta>0$ we found $x_1,x_2\in E$ with $|x_1-x_2|<\delta$ but $|f(x_2)-f(x_1)|\geq |f(x_1)|-|f(x_2)|>1.$

Hence $f$ is NOT uniformly continuous on $E$.

$\textit{Remark:}$ The condition that $x_0$ is a limit point of $E$ should be used when we want to prove that $f$ is unbounded in any nbhd of $x_0$.

Answer 2

I do not understand your proof, but I think it is easier to just use sequences. Since $x_0$ is a limit point of $E$, there is a sequence $(x_n)$ such that $x_n\to x_0.$ It is no loss of generality to assume $|x_n-x_0|<1$ for all integers $n$. Then, with $\epsilon=1,$ if $\delta>0$ is given, there is an integer $N$ such that $|x_N-x_0|<\min\{1,\delta\}$ and then of course $\frac{1}{|x_N-x_0|}>1,$ which contradicts the definition of uniform continuity.

Answer 3

Since $x_0\not\in E$ is a limit point of $E$, for any $\epsilon\gt0$, there is an $x_\epsilon\in E$ so that $|x_\epsilon-x_0|\lt\epsilon$.

Now if $f(x)=\frac1{x-x_0}$ were uniformly continuous on $E$, we should be able to find a $\delta\gt0$ so that if $|x-y|\lt\delta$, then $|f(x)-f(y)|\lt1$.

However, for any $\delta\gt0$, by the definition of $x_\epsilon$, $$ r=|x_{\delta/2}-x_0|\lt\delta/2\tag1 $$ By $(1)$, $$ |f(x_\delta)|=\frac1r\tag2 $$ Consider the point $x_{r/(r+1)}$. By the definition of $x_\epsilon$ and $(1)$, $$ \left|x_{r/(r+1)}-x_0\right|\lt\frac{r}{r+1}\lt r\lt\delta/2\tag3 $$ and so $$ \left|f\!\left(x_{r/(r+1)}\right)\right|\gt1+\frac1r\tag4 $$ By the Triangle inequality and $(1)$ and $(3)$, $$ \begin{align} \left|x_{\delta/2}-x_{r/(r+1)}\right| &\le\overbrace{|x_{\delta/2}-x_0|\vphantom{\left|x_{r/(r+1)}\right|}}^{\lt\delta/2}+\overbrace{\left|x_{r/(r+1)}-x_0\right|}^{\lt\delta/2}\tag{5a}\\[6pt] &\lt\delta\tag{5b} \end{align} $$ However, by the Triangle inequality and $(2)$ and $(4)$, $$ \begin{align} \left|f\!\left(x_{r/(r+1)}\right)-f(x_{\delta/2})\right| &\gt\overbrace{\left|f\!\left(x_{r/(r+1)}\right)\right|}^{\gt1+\frac1r}-\overbrace{\left|f(x_{\delta/2})\right|\vphantom{\left(x_{r/(r+1)}\right)}}^{\frac1r}\tag{6a}\\[6pt] &\gt1\tag{6b} \end{align} $$ Since $\delta\gt0$ was arbitrary, $(5)$ and $(6)$ show that $f$ is not uniformly continuous.