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If $E $ is a connected subset of $\Bbb R^{n} $, does a continuous mapping $f(E)$ into $\Bbb R^{m}$ preserve connectedness?

I think yes by definition of continuity since otherwise there would be some point s.t. $f(E)$ would be a disconnected set, but this point would be a discontinuity

Phillip Hamilton
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    This is somewhat easier to prove using the "topologists' definition" of continuity, and slightly harder using the "analyst's definition". For concreteness, if you edit your question to include what you know as the definition for continuity and connectedness, you will get an answer that best use those concepts. – Willie Wong Nov 17 '16 at 18:29

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Hint: Startby assuming $f: M \to N$ is continuous and surjetive and $M$ is connected. Suppose $N = A \cup B$, with $A,B$ disjoint open subsets of $N$. Now since $M$ is connected, what can you say about $f^{-1}(A)$ or $f^{-1}(B)$ where $M = f^{-1} (A) \cup f^{-1} (B)$?

Conclude that, for any $E \subseteq M$ connected, $f: E \to f(E)$, takes $E$ onto a connected subset of $N$.

Aaron Maroja
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The answer is yes and holds for the more general case that $E$ is any connected space. We prove this by proving the equivalent (contrapositive):

Let $f: X \rightarrow Y$ be continuous. If $f(X)$ is disconnected, then $X$ is disconnected.

Suppose that $f(X) = A \cup B$ is a separation, i.e. $A, B \subset f(X)$ is open, $A, B \neq \varnothing, A \cap B = \varnothing, A \cup B = f(X)$.

Let $C = f^{-1}(A), D = f^{-1}(B)$. Then $C, D \neq \varnothing, C \cap D = \varnothing$.

Write $A = f(X) \cap U$, where $U \subset Y$ is open.

Then $C = f^{-1}(A) = f^{-1}(U)$ is open, since $f$ is continuous.

Similarly, $D$ is open, and hence $X$ is disconnected.

Irregular User
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Suppose $\;f(E)\;$ isn't connected $\;\iff\;$ there exists a continuous onto map $\;h: f(E)\to\{0,1\}\;$ , the last two-elements set with the inherited euclidean topology from $\;\Bbb R\;$ (and thus discrete), but then:

$$E=(h\circ f)^{-1}(\{0\})\cup(h\circ f)^{-1}(\{1\})$$

and since both $\;(h\circ f)^{-1}(\{0\})\;,\;\;(h\circ f)^{-1}(\{1\})\;$ are open and non empty (why and why?), $\;E\;$ isn't connected.

DonAntonio
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  • The answer to your question is that a function maps an open set to an open set (or closed to closed), correct? – Phillip Hamilton Nov 17 '16 at 18:49
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    @PhillipHamilton No, that isn't true in general even in the case the function is continuous... What is true is that a continuous function's inverse image of an open (closed) set is open (closed), and $;h\circ f;$ is continuous and $;{0};,;;{1};$ are open and closed in $;{0,1};$ . – DonAntonio Nov 17 '16 at 19:54