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Prove $f: \mathbb{R} - \{0\} \to \mathbb{R}$ such that $f(x) = \frac{1}{x}$ is continuous using the topological definition of continuity

To do this all I need to do is pick a $U$ open in $(\mathbb{R} -\{0\}, d)$ and show that $f^{-1}[U]$ is open in $\mathbb{R}$.

$U$ is of the form $U =B_{(\mathbb{R} - \{0\}, d)}(f(x), \epsilon) = (f(x) - \epsilon), f(x) + \epsilon)$, and $f^{-1}[U] = \left\{x \ | \ f(x) = \frac{1}{x} \in U\right\}$. But I'm having trouble showing that $f^{-1}[U]$ is open in $(\mathbb{R}, d)$. Certainly we have $x \in f^{-1}[U]$ and from that I would have to find a $\delta > 0$ such that $f^{-1}[U] = B_{(\mathbb{R}, d)}(x, \delta)$.

Would it suffice to say that $f^{-1}[U] = \left(\frac{1}{f(x) - \epsilon)}, \frac{1}{f(x) + \epsilon)}\right)$ which is trivially open in $\mathbb{R}$?

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    $f(x)= \frac1x$ is not $\Bbb R\to \Bbb R$. It's $\Bbb R\setminus\{0\}\to \Bbb R$. – 2017-02-16
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    It would suffice, but a preimage of an interval must not be an interval! Consider $f^{-1}[(-1,1)]$ which is the union of $(-\infty, -1)$ and $(1,\infty)$. Consider showing the preimage is a union of open intervals. – 2017-02-16

3 Answers 3

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All you need is to prove that any basic open set of $\mathbb{R}$ (interval) has open pre-image, so ew have three cases for $U=]a,b[$: $a\geq0,b>0;a<0,b>0;a<0,b\leq0$

Is easy to see from $\mathrm{graph}f$ that:

first and third case: $f^{-1}(U)=]\frac{1}{b},\frac{1}{a}[$

second case: $f^{-1}(U)=]-\infty,\frac{1}{a}[\cup]\frac{1}{b},+\infty[$

Notice that in case $a=0$, $b=0$ we'll replace $\frac{1}{a}\rightarrow+\infty,\frac{1}{b}\rightarrow-\infty$ respectively.

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Let $U$ be open in ${\mathbb R}$, and consider an arbitrary point $a\in f^{-1}(U)$. We have to show that $f^{-1}(U)$ contains a full neighborhood of the point $a$. For simplicity assume $a>0$.

The point $b:={1\over a}>0$ is in $U$. Since $U$ is open there are numbers $u$, $v$ with $0 \subset U$. By the properties of the reciprocal function one has $$\left]{1\over v},{1\over u}\right[\>=f^{-1}\bigl(]u,v[\bigr)\subset f^{-1}(U)\ .$$ On the other hand ${1\over v}<{1\over b}<{1\over u}$, hence $\left]{1\over v},{1\over u}\right[\>$ is a neighborhood of $a={1\over b}$.

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Otherwise, let $x_0 \in \mathbb{R}\setminus \{0\}$ and $\varepsilon > 0$. Let $\delta =\frac{\varepsilon|x_0|^2}{1 + \varepsilon |x_0|}$. Then one has: \begin{align} |\frac{1}{x} - \frac{1}{x_0}| = |\frac{x_0 -x}{x_0x}|&\leq \frac{\delta}{|x_0|(|x_0| - \delta)}\\ &\leq \frac{\frac{\varepsilon|x_0|^2}{1 + \varepsilon |x_0|}}{|x_0|(|x_0| - \frac{\varepsilon|x_0|^2}{1 + \varepsilon |x_0|})}\\ &\leq \frac{\frac{\varepsilon|x_0|^2}{1 + \varepsilon |x_0|}}{ \frac{|x_0|^2}{1 + \varepsilon |x_0|}}\\ &\leq \varepsilon \end{align}

Then for any neighborhood $W$ of $f(x_0)$, there exists a neighborhood $W'$ in $\mathbb{R}^\star$ of $x_0$ such that $f(W') \subset W$. Indeed, given a neighborhood $W$, one can find $\varepsilon > 0$ such that $B(f(x_0), \varepsilon) \subset W$. Then let $W' = B(x_0, \delta)$ with $\delta$ as defined above. Computation above yields that $f(W') \subset B(f(x_0), \varepsilon) \subset W$, which proves continuity.