Continuity through the lens of the uniform norm

Question

So I was refreshing my knowledge of uniform convergence in the space of $C(X)$, continuous functions on a metric space, and wanted to ask whether the following characterization of continuity at a point $x\in X$ was equivalent to the typical metric space definition. Specifically, $f:X\to Y$ is continuous at $x$ if for all $\epsilon>0$ there eixsts $\delta >0$ such that: $$ \sup_{y \in B_\delta(x)} d(f(x),f(y)<\epsilon $$

This would also explain to me the reason the uniform norm is necessary to make $C(X)$ a complete metric space.

Answer 1

Let $\left(X, d_X\right)$ and $\left(Y, d_Y\right)$ be metric spaces, and let $f \colon X \to Y$ be a function. Let $p$ be a point of $X$. Then we give the following two definitions:

Definition 1

The function $f$ is said to be continuous at point $p$ if, for every real number $\varepsilon > 0$, we can find a real number $\delta = \delta(p, \varepsilon)$ (i.e. this real number $\delta$ may depend on our point $p$ and on our choice of $\varepsilon$) such that $\delta > 0$ and such that $$ d_Y \big( f(x), f(p) \big) < \varepsilon $$ for every point $x \in X$ for which $$ d_X (x, p) < \delta. $$

Definition 2

The function $f$ is said to be "specially continuous" at point $p$ if, for every real number $\varepsilon > 0$, we can find a real number $\delta = \delta(p, \varepsilon)$ (i.e. this real number $\delta$ may depend on our point $p$ and on our choice of $\varepsilon$) such that $\delta > 0$ and such that $$ \sup \left\{ \ d_Y \big( f(x), f(p) \big) \ \colon \ x \in B_{d_X} ( p; \delta ) \ \right\} < \varepsilon, $$ where $$ B_{d_X} ( p; \delta ) \colon= \left\{ \ v \in X \ \colon \ d_X(v, p) < \delta \ \right\}. $$

Now we prove that $f$ is continuous at $p$ if and only if $f$ is specially continuous at $p$.

Proof

Suppose that $f$ is continuous at $p$. Let $\varepsilon$ be a given real number such that $\varepsilon > 0$. Then, for the real number $\frac{\varepsilon}{2} > 0$, we can (by Definition 1) find a real number $\delta > 0$, depending on point $p$ and on $\varepsilon/2$, which in turn of course depends on $\varepsilon$, such that $$ d_Y \big( f(x), f(p) \big) < \frac{\varepsilon}{2} $$ for every point $x \in X$ for which $$ d_X (x, p) < \delta; $$ that is, the set $$ \left\{ \ d_Y \big( f(x), f(p) \big) \ \colon \ x \in B_{d_X} (p; \delta ) \ \right\} \tag{1} $$ is a non-empty and bounded above subset of $\mathbb{R}$ and $$ \sup \left\{ \ d_Y \big( f(x), f(p) \big) \ \colon \ x \in B_{d_X} (p; \delta ) \ \right\} \leq \frac{\varepsilon}{2}. $$ Hence $$ \sup \left\{ \ d_Y \big( f(x), f(p) \big) \ \colon \ x \in B_{d_X} (p; \delta ) \ \right\} < \varepsilon. $$ Since $\varepsilon > 0$ was arbitrary, it follows that $f$ is specially continuous at $p$.

Note that the set in (1) above is non-empty because the point $p$ is always in the set $B_{d_X}(p;\delta)$ since $$ d_X(p, p) = 0 < \delta. $$

The converse is even easier to prove. Hope you can do that yourself.