Proving that the component spaces are connected if the product space is connected

connectednessfunctionsgeneral-topologyopen-mapproof-writing

Show that if $\prod_\alpha X_\alpha$ is connected and nonempty then each $X_\alpha$ is connected.

My work so far involves using this property of a connected space: the only open and closed sets are $\emptyset$ and the space itself.

I assume to know that the canonical projection maps $p_\alpha : \prod_\alpha X_\alpha \to X_\alpha$ are open maps.

First I consider $U=\emptyset\in \prod_\alpha X_\alpha$. Then $U$ is open, so $p_\alpha (U)$ is also open, and its complement $\emptyset^c = X_\alpha$ will be closed. And vice versa if $U=\prod_\alpha X_\alpha$ then $p_\alpha(U)=X_\alpha \implies X_\alpha^c =\emptyset\in X_\alpha$ is closed. So we get that $\emptyset$ and $X_\alpha$ are clopen in each $X_\alpha$.

Next is the case that $V$ is a proper subset of $\prod_\alpha X_\alpha$. Then I know $V$ is either open or closed, but not both.

This is where I get stuck, since the projection maps are open, but not necessarily closed, so I know if $V$ is open then $p_\alpha (V)$ is also open, but there's no reason I can see so far why $p_\alpha (V)$ is not also closed in each $X_\alpha$.

Is there a way to continue this line of reasoning to complete the proof?

Best Answer

You were good until (as pointed in the comments ) you assumed that $V$ has to be either open or closed which is not true (it can be neither)

The simplest example I can think of is $[0,1)$ the half open interval it is not closed because $1$ which is the limit point of $[0,1)$ is not contained in our set which by definition should be contained in our set if it were to be closed. Another example includes $\mathbb{Q}$ (try to prove this yourself using the fact that there is a sequence in $\mathbb{Q}$ which converges to a irrational number and consider the $\epsilon$-neighborhoods for the not open case) $■$

Now to prove the question generally we would be looking at the projection maps

Proposition:A projection map denoted $\pi_{\alpha} : \prod_{\alpha}X_{\alpha} \rightarrow X_{\alpha}$ is continuous.
Proof : The basis of the topology $ \prod X_{\alpha} =$ {$ \prod_j U_{\alpha} \mid U_{\alpha} \in X_{\alpha}$}. Now let $V$ be open in $X_{\alpha}$, $\pi_{\alpha}^{-1}(V) = V \times \prod_{\beta \neq \alpha}$ which is open in the product topology where $\beta$ and $\alpha$ are in the same index set (which essentially says all the $X_i$ except $X_{\alpha}$ in the product topology). $□$

Now notice that these projections are canonically surjective which you can check follows from the definition: $f: X \rightarrow Y$ is said to be surjective if for every $y$ in $Y$ there exists some $x$ in $X$ such that $f(x) = y$ (basically the if $f$ is surjective then the codomain $=$ domain.)

Now we are essentially done because continuous functions preserve connectedness

Lemma: Continuous (surjective) functions preserve connectedness.
Proof: Let $f: X \rightarrow Y$ be surjective and let $X$ be connected. Now for the sake of contradiction suppose that there are some disjoint $A$ and $B$ such that $A \cup B = Y$. Now $f^{-1}(A)$ and $f^{-1}(B)$ are disjoint and since $f$ is surjective they form a separation of $X$ which contradicts the fact that $X$ is connected hence $Y$ must also be connected. The other way around also works and follows a very similar stream of thought. Let $Y$ be connected and assume for the sake of contradiction that $x$ is not connected. Then there exists some $A,B$ such that $A \cap B = \phi$ (disjoint) and $A \cup B = X$. Now since $f$ is continous and surjective $f(A)$ and $f(B)$ form a separation of $Y$ which contradicts the fact that $Y$ is connected $□$

Extras/References
This is something new (this is the first!) I want to try out where I put forth a few (related) questions and maybe references to articles or books and personal notes if needed

  1. Prove that $\mathbb{Q}$ is neither closed nor open

  2. Prove that the projection map $\pi_{\alpha}$ is continuous for infinite products

  3. In the Lemma we we said or assumed that $f$ is surjective. What if we do not assume $f$ to be surjective would connectedness still be preserved?

  4. What other properties are preserved by continuous functions? (Like we just proved connectedness in preserved are other properties like compactness locally-compact/connected or even the separation axioms preserved too?)

  5. To put forth some references Engelking has a great book (with imo quite hard problems) others inlude John Kelley I would also recommend Counterexamples in topology (Dover edition)

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