I started studying topology and encountered the epsilon-delta definition of continuity applied for general metric spaces. From my calculus courses I am used to thinking of both $\epsilon$ and $\delta$ in the epsilon-delta definition of continuity as limits tending to zero. That is (as I understand now) is equivalent to the statement that a continuous function maps every converging Cauchy sequence from its domain into a converging Cauchy sequence in its range.

However what if there are no Cauchy sequences in some metric space (or they just do not converge in this space)? Does it mean that on such domain no continuous functions can be defined?

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    $\begingroup$ Every point in metric space has some Cauchy sequence which converges to that point - consider the constant sequence. $\endgroup$ – Hanul Jeon Nov 8 '14 at 0:59

If a metric space has no nontrivial Cauchy sequences that converge to a point in the space, then it has the discrete topology, so every function is continuous. We can see this by noting that for each point $x$ there is a number $\epsilon_x>0$ such that $d(x,y)\geq \epsilon_x$ for all $y\neq x$ (for otherwise we could construct a Cauchy sequence that converges to $x$). Therefore the open ball $B_{\epsilon_x/2}(x)$ consists of the single point $x$, so $\{x\}$ is open.

  • $\begingroup$ Ok, thank you. But another question then on the discrete topology. The epsilon-delta criteria reads $\| x - c | < \delta \Rightarrow | f(x) - f(c) | < \epsilon$, with $\x$ and $\c\$ being two distinct points. Does it mean that in a discrete topology $\delta\$ and $\epsilon\$ are not tending to zero? Does not it mean then the $\delta-epsilon\$ criteria fails here? Or am I missing something here? $\endgroup$ – Dv139 Nov 8 '14 at 3:43
  • $\begingroup$ Sorry the edit button is gone :( $\endgroup$ – Dv139 Nov 8 '14 at 3:58
  • $\begingroup$ It actually makes the criterion easier to satisfy. If there are no points within $\delta$ of $x$ then $|x-c|<\delta$ is false, so any implication depending on that is true. $\endgroup$ – Matt Samuel Nov 8 '14 at 4:10
  • $\begingroup$ false or undefined? I am also trying to comfort my calculus rationale (which is terribly wrong sometimes) that the derivative of a function with domain in discrete topology makes little sense. And derivative is directly linked to the epsilon-delta criteria as I understand. $\endgroup$ – Dv139 Nov 8 '14 at 4:23
  • $\begingroup$ Derivatives are not required for epsilon delta proofs. We are dealing with continuity here, not differentiability. $|x-c|<\delta$ is defined and false for all $c$ for small enough delta at such a discrete point. Therefore $|x-c|<\delta|\Rightarrow|f(x)-f(c)|<\epsilon$ is vacuously true. $\endgroup$ – Matt Samuel Nov 8 '14 at 12:23

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