Is this alternative notion of continuity in metric spaces weaker than, or equivalent to the usual one?

Solution 1:

Indeeed definition $2$ implies definition $1$:

Let us assume that $f$ satisfies definition $2$, and try to prove it is continuous by definition $1$.

Let $\epsilon > 0$. Then, because $\lim_{n\to\infty} \epsilon_n = 0$, there exists such a $N$ that $\epsilon_N < \epsilon$.

Now, let us set $\delta=\delta_N$, and let $|x-x_0|<\delta$. Because $\delta=\delta_N$, this means that $|x-x_0|<\delta_N$, and by definition $2$, this means that $|f(x)-f(x_0)|<\epsilon_N$.

This further means that $|f(x)-f(x_0)|<\epsilon_N<\epsilon$, meaning that $f$ is continuous at $x_0$.

Solution 2:

The second definition may not require that a $\delta$ exists for every $\epsilon > 0$, but we still get this as a consequence. This is because if $\delta$ works for a given $\epsilon$, then $\delta$ will work for any larger $\epsilon$, too. So, $\delta_{n}$ will work for every $\epsilon \geq \epsilon_{n}$. Since $\epsilon_{n} \to 0$, then your second "definition of continuity" implies the original.