Is it true that the intersection of a sequence $K_1 \supset K_2 \supset K_3 \dotsm$ of connected subsets of $\mathbb{R}^2$ is also connected?

Question

I have got one counterexample for this : Consider the family {D} of closed discs centered at zero having radius $1+1/n$, i.e. disc $D_1$ has radius $1+1=2$, $D_2$ has radius $1+1/2=1.5$, and so on. Now consider the family {D'} of sets of the form $D_i-\{(0,y)\mid -1 \leq y \leq 1 \}$. Each of these sets is connected but their arbitrary intersection is not. Is this example is correct ? If not please provide another one.

possible duplicate of Prove that $\bigcap_{k = 1}^\infty C_k$ is also compact and connected. — Surb, May 16 '15 at 11:40
@DavidMitra this is correct but the link in question is a duplicate of another link that does not deal with compactness — Surb, May 16 '15 at 11:44
This question does not deal with compactness. example which I have mentioned have sets of the form neither open nor close. and since we are considering usual topology of R2, only closed and bounded sets can be considered as compact. — Math_Explorer, May 16 '15 at 11:59

score 4 · Accepted Answer · answered May 16 '15 at 12:15

Here is another example involving closed (but not bounded) sets. Let $A_n = (-1, 1) \times (-n, n)$. Let $B_n = \mathbb R^2 - A_n$. Each $B_n$ is connected. We have $$ \bigcap_n B_n = ((-\infty, -1] \times \mathbb R) \cup ([1, \infty) \times \mathbb R). $$

Thus, $\bigcap_n B_n$ is disconnected.

Is it true that the intersection of a sequence $K_1 \supset K_2 \supset K_3 \dotsm$ of connected subsets of $\mathbb{R}^2$ is also connected?

1 Answers1