0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

Globally, it is fine. Of course, you should have told us what are the contents of exercise 3.4.2, but that's easy to guess.

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 $V,W\neq\emptyset$, $f^{-1}(V),f^{-1}(W)\neq\emptyset$.

$\endgroup$
1
$\begingroup$

The idea of the proof is fine. But to simplify this, get rid of all the "relative" stuff.

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 $X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.