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Exercise:

In the definition of continuity show that the centered intervals $$|f(x) - f(x_0)| < \epsilon \text{ and } |x-x_0| < \delta$$ may be replaced by an arbitrary open interval containing $f(x_0)$ and a sufficiently small open interval containing $x_0$.

Attempt:

$|f(x)-f(x_0)| < \epsilon$

$f(x) - \epsilon < f(x_0) < f(x) + \epsilon$

Let $f(x_0) > \epsilon_{\text{lo}} > f(x) - \epsilon$ and $f(x) + \epsilon > \epsilon_{\text{hi}} > f(x_0)$

$\epsilon_{\text{lo}} < f(x_0) < \epsilon_{\text{hi}}$

Take similar steps with $|x-x_0| < \delta$

$\square$

Is my attempt correct? If not, where and why'd I go wrong?

  • 0
    Try to show the more general "for all open sets $U$ in the range, the pre-image $f^{-1}(U)$ is open". Your idea may be right in the end, but it is more complicated than necessary. – 2017-01-12

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I think you have the right idea in your mind. However, it should be phrased more carefully. Let's redefine continuity at $x_0$.

$\forall \epsilon_1 , \epsilon_2>0 \exists \delta_1 , \delta_2>0 \hspace{0.1cm}| \hspace{0.1cm} x_0-\delta_1

I suppose, if we can prove this for a function $f(x)$, then we can readily have the regular definition of continuity. If we can have the above definition satisfied for a function $f(x)$, for any two $\epsilon_1$ and $\epsilon_2$, then we may take the special case $\epsilon_1=\epsilon_2$. For them, we get $\delta_1$ and $\delta_2$ that satisfy

$x_0-\delta_1

It remains to take $\delta=min(\delta_1,\delta_2)$