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The problem is more formally as follows: Let $A\subset \mathbb{R}$ and let $f:A \to \mathbb{R}$ be continuous at $c \in A$. Show that $\forall \epsilon >0$, there exists a neigbhorhood $V_\delta (c)$ of $c$ such that if $x,y \in A \cap V_\delta (c)$, then $\left|f(x) - f(y) \right| < \epsilon$.

My attempt...

Proof

By hypothesis, $\forall \epsilon > 0 \, , \exists \delta > 0 $ such that $\forall x,y$ satisfying $\left|x - c\right| < \delta$, $\left|y - c\right| < \delta$ it holds that

$$\left|f(x) - f(c) \right| < \epsilon/2 \\ \left|f(y) - f(c) \right| < \epsilon/2$$

So by the triangle inequality

$$ \left|f(x) - f(y)\right| \leq \left|f(x) - f(c)\right| + \left|f(c) - f(y) \right| < \epsilon$$

For some reason I'm getting the feeling that this isn't rigorous enough. Futhermore, as the title implies I've taken this fact to mean that if a function is continuous at a point that it is also uniform continuous in some small neigborhood of that point; is that correct? In this problem $x,y$ are within 2$\delta$ of each other $\implies \left|f(x) - f(y)\right| < \epsilon$.

Zduff
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    You will not be able to find a valid argument since the statement is not true. I linked an example of a function which is continuous at exactly one point, and thus not continuous (nor uniformly continuous) on any interval. https://math.stackexchange.com/questions/60050/find-a-function-f-mathbbr-to-mathbbr-that-is-continuous-at-precisely-o – Quoka Feb 03 '18 at 23:52

1 Answers1

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Your proof is correct for the statement from the body of your question.

However, as people have pointed in the comments, this is not the same statement as in the title. The condition in the body does not imply the condition in the title. For the condition in the title, $\delta$ must not depend on $\varepsilon$.

Counterexample of the statement in the title: Set $f(x)=xI_{\mathbb Q}(x)=\begin{cases}x & x\text{ rational}\\0 & x\text{ irrational}\end{cases}$, where $I_{\mathbb Q}$ is the Dirichlet function. It can be proven that it is continuous only in the point $x=0$, so it cannot be uniformly continuous on any neighbourhood of $0$.