# Continuity preserves connectedness

I am self-studying mathematics without a mentor, which is why I am interested in your opinion in the following proof. How would you rate it as a professor or PhD-Student at a university? As an undergraduate math major, do you get to submit such proofs in Analysis classes? Or are they considered too trivial for a homework?

Definition 2.4.1 (Connected spaces). Let $(X, d)$ be a metric space. We say that X is disconnected iff there exist disjoint non-empty open sets V and W in X such that $V \cup W = X$. We say that $X$ is connected iff it is non-empty and not disconnected.

Theorem 2.4.6. (Continuity preserves connectedness) Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Let $f:X\to Y$ be a function, $f$ is continuous. Let $E \subseteq X: E$ is connected. Then $f(E)$ is also connected.

Proof: Suppose, for the sake of contradiction, that $f(E)$ is not connected, i.e., there exist

• non-empty
• disjoint
• open relative to $f(E)$

sets $V$ and $W$ which cover $f(E) = V \cup W$. Now consider the induced metric spaces $(E,d_X|_{E\times E})$ and $(f(E),d_y|_{f(E) \times f(E)})$ and the function $$f|_{E}: E \to f(E), \quad f|_{E} (x) := f(x)$$ Obviously, $f|_{E}$ is surjective; hence, due to the Exercise 3.4.2, $$E = f|_{E}^{-1} \left( f|_{E} (E) \right)$$ Also, $f(E) = f|_{E} (E)$, hence, $$E = f|_{E}^{-1} \left( f (E) \right)$$ By the assumption of contradiction $$E = f|_{E}^{-1} \left( V \cup W \right)$$ Due to the properties of a inverse function (Exercise 3.4.2) we can rewrite this expression as $$E = f|_{E}^{-1} \left( V \right) \cup f|_{E}^{-1} \left( W \right)$$ But then we face the fact that both $f|_{E}^{-1} \left( V \right)$ and $f|_{E}^{-1} \left( W \right)$ are

• open relative to $E$,
since restriction of a continuous function preserves continuity; hence, $f|_{E}$ is continuous and preserves openness of an inverse image;
• non-empty,
since $V$ and $W$ are non-empty, and $f|_{E}$ is surjective
• disjoint,
by definition of function.

Thus, we have found two non-empty, disjoint, open relative to $E$ sets $f|_{E}^{-1} \left( V \right)$ and $f|_{E}^{-1} \left( W \right)$ such that $E = f|_{E}^{-1} \left( V \right) \cup f|_{E}^{-1} \left( W \right)$ - a contradiction to the connectedness assumption.

The only point about which a have a problem consists in the assertion that “$f|_E$ is surjective”. No, it is not, at least not in general. What happens is that you are assuming that both $V$ and $W$ are subsets of $f(E)$. Therefore, since $V,W\neq\emptyset$, $f^{-1}(V),f^{-1}(W)\neq\emptyset$.
If $f: X \to Y$ is continuous, so is $f: E \to f[E]$. (restrict the map on domain and co-domain).
So it suffices to show that $f:X \to Y$ continuous and onto, $X$ connected, then $Y$ is connected. And this is a matter of noting that when $\{V,W\}$ is a decomposition of $Y$, then $\{f^{-1}[V], f^{-1}[W]\}$ is one for $X$.