Implied By/Implies Continuity Using the Definition of Continuity

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I have to determine which of some definitions are implied by or imply the continuity of a function in a point.

So I guess I should play with $\varepsilon$ and $\delta$ one time I understood what the definition means (using a graph for example).

Definition $\mathcal A$

$$\forall \varepsilon^* \gt 0 \quad \exists \delta^* \gt 0 : \quad if \quad |x-x_0|\lt \varepsilon^* \Rightarrow |f(x)-f(x_0)| \lt \delta^*$$

I tried to draw some graphs and it seems to me that in this case the definition of continuity implies the definition $\mathcal A$

Then I tried to prove that the continuity implies $\mathcal A$

We have (definition of continuity)

$$\forall \varepsilon \gt 0 \quad \exists \delta \gt 0 : \quad if \quad |x-x_0|\lt \delta \Rightarrow |f(x)-f(x_0)| \lt \varepsilon$$

Then we have that there exists a $\delta$ for the definition of continuity: I pick this $\delta$ and I use it inside $\mathcal A$

So I choose $\varepsilon^* = \delta $, for that $\delta$ I have that $|x-x_0|\lt\delta$, then $|x-x_0|\lt\varepsilon^*$

Before stating that $\delta$ existed I had to pick an $\varepsilon$, so now I can say there is a $\varepsilon$ such that $|f(x)-f(x_0)| \lt \varepsilon$

I choose $\delta^* = \varepsilon$, then I have

$$\forall \varepsilon^* \gt 0 \quad \exists \delta^* \gt 0 : \quad if \quad |x-x_0|\lt \varepsilon^* \Rightarrow |f(x)-f(x_0)| \lt \delta^*$$

The only thing that makes me unsure that this is a correct proof is the fact that I state $\forall \varepsilon^* = \delta$

But not $\forall \varepsilon^* \gt 0$

Does that make sense?

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1 Answer

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Note that your definition $\cal A$ can be interpreted as following:

Given any $\epsilon^* > 0$, the function $f$ is bounded on $(x_0 - \epsilon^*, x_0 + \epsilon^*)$. If your function is defined on $\Bbb R$, then continuity does imply that, as can be seen by the Extreme Value Theorem.

However, it is not true in general. For example, consider $f : (0, 1) \to \Bbb R$ defined by $$f(x) = \dfrac{1}{x}.$$

Consider $x_0 = \dfrac{1}{2}$ and $\epsilon^* = \dfrac{1}{2}$. You should be able to show that no $\delta^* > 0$ exists.

(Note that $$\sup_{|x - x_0| < \frac12} |f(x) - f(x_0)| = \infty$$

in this case.)

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