In one of my Calculus exercises sets, I have to show that:

For an indexed family of connected sets $(X)_a$ with a non-empty intersection, their union is also connected.

My attempt :

Since each $X_a$ is connected, every continuous function $f:X_a \rightarrow \{0,1\}$ is constant. Now take $a \in \cap X_a$, now for every $X_a$ the value of $f(X_a)$ is constant and is actually equal to $f(a)$. So if $x \in \cup X_a$ then it belongs to a certain $X_i$ such that $f(x)$ is equal to $f(a)$, by the previous argument. So $f:\cup X_a \rightarrow \{0,1\}$ is constant.

I think my proof is missing something and I'm not quite sure if it actually proves the statement. Any recommendations or comment are welcome.

  • 1
    $\begingroup$ It looks fine. What's your doubt? $\endgroup$ – José Carlos Santos Sep 19 at 14:56

Your proof is fine, though I'd structure the order differently:

Let $f: \bigcup_{a \in A} X_a \to \{0,1\}$ be continuous.

Pick $p \in \bigcap_{a \in A}$ which exists by assumption.

Now for any arbitrary $a \in A$: $f\restriction_{X_a}: X_a \to \{0,1\}$ is also continuous and hence constant, so $x \in X_a \implies f(x)=f(p)$. So for any $x \in \bigcup_{a \in A} X_a$ we also have $f(x)=f(p)$.

So $f$ is constant and so $\bigcup_{a \in A} X_a$ is connected, using the function characterisation.


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