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).

But the statement is true in general, not just for metric spaces, and it's a pretty easy consequence of the definition of continuous in a general topological space:

A function $f:X \rightarrow Y$ is continuous if $U \subseteq Y$ open in $Y\Rightarrow f^{-1}(U)$ open in $X$.

  • $\begingroup$ 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. $\endgroup$ – Not Legato Feb 14 at 23:07
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    $\begingroup$ You could try converting the general proof into the metric space context via an $\epsilon-\delta$ argument. $\endgroup$ – Robert Shore Feb 14 at 23:13
  • $\begingroup$ Made a mistake in my comment due to hastiness; the edit is wrong, and the proof cannot be fixed by phrasing the existence of r differently. $\endgroup$ – Not Legato Feb 15 at 4:56

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