I'm trying to demonstrate the connectedness of the set $A_j = \{(1,0),(0,0)\} \cup \{(x,y) : 0 < y < 1/j\}$. This is for my class in real analysis, so I can't apply concepts that are too specific to basic topology.

It seems very clear to me that is is indeed connected, but I'm not sure how formally state it in a proof. I was thinking that I could say that $\{(x,y) : 0 < y < 1/j\}$ is path-connected and therefore connected because if you take any two points, you can connect them with a straight line that is, by construction, completely enclosed within the set. What's a good way to argue that the union of this set with points on its boundary (namely, $(0,1)$ and $(0,0)$) is also connected?

I was thinking that I could argue that if $A$ is connected and $A \subset B \subset \overline{A}$, then $B$ is connected, and apply that to this problem. Indeed, I think that I'm able to prove this:

Suppose $(X,Y)$ is a disconnection of $B$. Then $A \subset X \cup Y$, and obviously the intersection of $A$ with at least one of $X,Y$ is nonempty. Suppose that both $A \cap X \ne \emptyset$ and $A \cap Y \ne \emptyset$. Then the intersections $X' = X \cap A$ and $Y' = Y \cap A$ are nonempty and $(X',Y')$ is a disconnection of $A$, which is a contradiction. Alternatively, suppose without loss of generality that $A \cap X \ne \emptyset$ but $A \cap Y = \emptyset$, \ie, $A \subset X$. Recall from the definition of a disconnection that $\overline{X} \cap Y = \emptyset$. Since $A \subset X$, it follows that $\overline{A} \subset \overline{X}$, and so $\overline{A} \cap Y = \emptyset$. Hence there is at least one element $y \in Y \subset B$ such that $y \ne \overline{A}$. But $B \subset \overline{A}$, so this is a contradiction.

However, is there an easier way?


Your proof is very nice, but it's far more general than what you need.

Suppose $(U,V)$ is a partition of $A$ into nonempty open sets. Let $$U' = U - \{ (0,0),(1,0) \} \quad \text{and} \quad V' = V - \{ (0,0),(1,0) \}$$ Then $U' \ne \varnothing$ and $V' \ne \varnothing$ since $(0,0)$ and $(1,0)$ lie in the closure of $\{ (x,y) : 0 < y < \frac{1}{j} \}$. Then $(U',V')$ is a partition of $\{ (x,y) : 0 < y < \frac{1}{j} \}$ into nonempty open sets, which is nonsense because this set is connected.

  • $\begingroup$ Oh, I see -- that's definitely a lot simpler, I can't believe I didn't think of that! Thanks! $\endgroup$ – Marcus Emilsson Oct 20 '14 at 1:05

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