$F= bigcap_{i=1}^{infty} F_i$ isn't necessarily connected where $F_{i+1} subseteq F_i$ and $F_i subseteq mathbb{R}^2$ are closed and connected

Question

In my attempt, I first show that $F$ is closed, this is since we can write $F= bigcap_{i=1}^{infty} F_i = (bigcup_{i=1}^{infty} F_i^C)^C$ and $bigcup_{i=1}^{infty} F_i^C$ is a union of open sets, hence it is open, hence $F=(bigcup_{i=1}^{infty} F_i^C)^C$ must be closed.
From here, I am trying to construct an example. I was thinking of $F_i$ being two disjoint disks that are connected with a thick line between them, such that as $i$ grows the line becomes thinner and its thickness tends to zero. The problem is that in this case $F$ still has a one-point width line connecting the disks and this seems not good enough.
Will be happy for any help on this or a better example (Assuming the standart topology on $mathbb{R}^2$)

Eric Towers · Answer

Another example, $F_i$ us the union of the closed vertical rays starting at $(0,0)$ and $(1,0)$ and the closed half-plane above abscissa $i$.
$$  F_i = {(0,y) : y geq 0} cup {(1,y) : y geq 0} cup {(x,y) : y geq i}  $$
As $i$ increases, the "bridge" between the two rays is yanked away.

MPW · Answer

Consider taking $F_i$ as the closed upper half-plane with $(0,1)times [0,i)$ removed.

$F= bigcap_{i=1}^{infty} F_i$ isn't necessarily connected where $F_{i+1} subseteq F_i$ and $F_i subseteq mathbb{R}^2$ are closed and connected

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