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I'm trying to prove that, if f is a function from C to C, and its domain, D, is connected, then f(D) is also connected. How would I go about doing this?

The definition of conectedness at play is "S is disconnected iff there exist open disjoint sets A and B such that none contains S, but their union does", and that of continuity is "f is continuous iff, if a sequence of members of the domain tends to z, then the image of the sequence tends to f(z)".

Robly18
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    Do you know the definition of a continuous function and a connected set? – Daniel McLaury Nov 17 '16 at 19:10
  • Suppose there is a continuous function $g:f(D)\to\Bbb R$ with image ${0,1}$, then $g\circ f$ is a continuous function $D\to\Bbb R$ with image ${0,1}$. –  Nov 17 '16 at 19:11
  • If $f$ is continuous, then inverse images of open sets are open. Suppose $f(D)$ is not connected, then show that $D$ is not connected. – copper.hat Nov 17 '16 at 19:20
  • @G.Sass Interesting proof. I would guess that most texts don't use this as the definition of connectedness, so you'd also have to prove that if a set is disconnected then such a function into $\mathbb R$ exists (which is not difficult in any case). Maybe you'd like to post your alternative solution as an answer. – Matt Samuel Nov 17 '16 at 20:12
  • @MattSamuel someone already did. –  Nov 17 '16 at 20:13

2 Answers2

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Assume $f(D)\subset U\cup V$ where $U,V$ are open sets in $\Bbb{C}$. $f$ being continuous the inverse image of an open set is an open set and so $f^{-1}(U),\,f^{-1}(V)$ are therefore open sets in $\Bbb{C}$.

Now take $x\in D$, one has $f(x)\in f(D)$ and this means $f(x)\in U$ or $f(x)\in V$ and so $x\in f^{-1}(U)\cup f^{-1}(V)$ and therefore $D\subset f^{-1}(U)\cup f^{-1}(V)$.

But $D$ is connected so it cannot be included in the union of two disjoint open sets and so $\exists a\in f^{-1}(U)\cap f^{-1}(V)$ and thus $f(a)\in U\cap V$. Therefore $U,V$ are not disjoint and $f(D)$ is connected.

mxs
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marwalix
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  • Thanks for the response. One question, however: what if, say, U has elements that are NOT in f(D)? Thus, would it make sense to consider the objects corresponding to U? – Robly18 Nov 17 '16 at 19:43
  • @Robly18 $U$ will contain some points not in $f(D)$ quite often. That's not a problem though. You can still take the inverse image. – Matt Samuel Nov 17 '16 at 19:47
  • @MattSamuel Can you, despite this, guarantee that the inverse image is open? After all, consider the identity function with domain the interior and edge of the unit circle. Then, let U be the interior of the radius 2 circle. What will the inverse image of f be in regard to U? – Robly18 Nov 17 '16 at 19:52
  • @Robly18 The inverse image of every open set is open, regardless of whether or not the function is surjective onto the open set. It's the definition of continuity. The inverse image of any open set is open. – Matt Samuel Nov 17 '16 at 20:00
  • Isn't in last paragraph of proof, $a$ must belongs to intersection of $f^{-1}(U)$ and $f^{-1}(V)$? – Akash Patalwanshi Apr 05 '18 at 16:06
  • A good answer, although it still requires some thinking. – Ziqi Fan Jan 08 '21 at 22:45
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Suppose that $g:f(D)\rightarrow\{0,1\}$ is continues, (where we equip $\{ 0, 1 \}$ with the discrete topology $\big\{ \emptyset, \{ 0 \}, \{ 1 \}, \{ 0, 1 \} \big\}$) then $g \circ f$ is continues. Since $D$ is connected, we deduce that $g\circ f$ is constant. This implies that $g$ is constant since the image of the restriction of $f$ to $D$ is $f(D)$. We deduce that $f(D)$ is connected.

ViktorStein
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Tsemo Aristide
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