# The image of a continuous mapping on a connected metric space is connected: ($\epsilon - \delta$) proof

I've seen entirely set-theoretical solutions to this proof, but not many based on epsilon-delta arguments. I tried writing one, but am not sure whether my arguments hold.

Theorem: The image of a continuous mapping on a connected metric space is connected.

Proof: Let $$\mathrm{(X,d_1)}$$, $$\mathrm{(Y,d_2)}$$ be metric spaces, where $$\mathrm X$$ is a connected space, and let $$f:\mathrm{X} \rightarrow \mathrm{Y}$$ be a continuous mapping.

Assume that $$f[\mathrm{X}] = \mathrm{P \cup G}$$, where $$\mathrm{P,G}$$ are nonempty, disjoint, open sets.

$$\mathrm{P}$$ and $$\mathrm{G}$$ are disjoint $$\Rightarrow \exists r>0$$ such that $$\mathrm{B(x,r)}\cap\mathrm{G} = \emptyset\space \forall x\in \mathrm{P}$$.

$$f$$ is continuous $$\Rightarrow$$ $$\forall \epsilon>0,\space\exists \delta>0$$ so that if $$d_1(x,y)<\delta \Rightarrow d_2(f(x),f(y))<\epsilon = \frac{r}{2}$$ for all $$x\in f^{-1}[\mathrm{P}], y\in f^{-1}[\mathrm G]$$

$$\Rightarrow \exists\epsilon < r \space\forall r>0 \Rightarrow\nexists \mathrm{B(x,r)}\cap\mathrm{G} = \emptyset$$, which contradicts the assumption that $$\mathrm{P,G}$$ are disjoint and open.

$$\Rightarrow \mathrm{P,G}$$ are closed and not disjoint.

Now, there are a few questions I have: I tried using a point inside the open ball such that the distance between the point and the set G was minimised, and tried to show that $$r < \epsilon$$, but couldn't make it work, no matter what I tried. Is it viable?

Also, I feel like the proof is a bit cluttered, like something could be cleaned up, and concluding that P,G must be closed and disjoint, I feel like it should be phrased differently for clarity.

I'm afraid your proof doesn't work. Let $$Y = \Bbb R$$ with the usual metric, let $$P$$ be the negative reals and let $$G$$ be the positive reals (so there's a "hole" at $$0$$ that disconnects $$P$$ and $$G$$). It's not true that there is some $$r$$ that uniformly separates $$P$$ and $$G$$ (in other words, there's no $$r$$ that works for all $$x \in P$$, though for each $$x \in P$$ there is an $$r$$ that works).
A function $$f:X \rightarrow Y$$ is continuous if $$U \subseteq Y$$ open in $$Y\Rightarrow f^{-1}(U)$$ open in $$X$$.
• Is there a reasonable way to prove the statement using an $\epsilon-\delta$ argument, then, for the sake of practice alone? Seems like this proof is unsalvageable without adding a constraint that $\mathrm{X}$ is closed, which is too specific already. Edit: I misread. So the proof could be fixed by phrasing the existence of the r differently? I actually meant for there to exist an r for some x, so that's great, easily fixed. – Not Legato Feb 14 at 23:07
• You could try converting the general proof into the metric space context via an $\epsilon-\delta$ argument. – Robert Shore Feb 14 at 23:13