Proposition. Let $(X,d)$ be a metric space and $A_i \subset X$, $i \in I$ where $I$ is some interval. If each $A_i$ is connected and $A_i \cap A_j \neq \emptyset$ for all $i,j \in I$ then $A = \bigcup_{i \in I} A_i$ is connected.

I've proved the proposition above and what to use it to show that open and semi-open intervals, i.e. $(a,b)$, $[b, \infty)$, and $[a, -\infty)$, are all connected in $\mathbb{R}$.

Is the following enough, or should I make it stronger:

For the sake of contradiction, suppose $(a,b)$, $[b, \infty)$, and $[a, -\infty)$ are all disconnected. Then they are disjoint, non-empty, and relatively open sets $U$ and $V$ such that either $$(a,b) = U \cup V ~~~~~~~~~~ [b, \infty) = U \cup V ~~~~~~~~~~ (-\infty, a] = U \cup V.$$ Choose some point $x \in U$ and $y \in V$ and construct the connected interval $I = [x,y]$ (one can assume that $[x,y]$ is connected in $\mathbb{R}$). But, by the proposition above, then we must have that $U \cap I$ and $V \cap I$ are disjoint, non-empty, and relatively open in $I$. Meaning, we have a contradiction since $I$ is connected.

  • 1
    $\begingroup$ your proof is correct, and you did not use the proposition at all. $\endgroup$
    – user99914
    Oct 6, 2017 at 4:31
  • 1
    $\begingroup$ Pedantically, because of the way the "For the sake of contradiction" sentence is phrased, you would only show that for all $a,b$, at least one of $(a,b),$ $[b,\infty),$ and $[a,-\infty)$ is connected (i.e. the negation of the statement that they are all disconnected). And using the same $U,V$ for each case does not make sense. $\endgroup$
    – Dap
    Oct 6, 2017 at 6:04
  • $\begingroup$ @Dap I see what you mean. However, breaking this into 3 separate cases we would just apply the same process, correct? Was trying to kill three birds with one stone... $\endgroup$
    – jj8989
    Oct 6, 2017 at 7:39
  • 1
    $\begingroup$ @jj8989: well you can either rephrase it as "we will treat all three cases in the same way" in leave the reader to check the argument works in each case, or find some sufficient property that all cases have in common (i.e. convexity) $\endgroup$
    – Dap
    Oct 6, 2017 at 13:45


You must log in to answer this question.

Browse other questions tagged .