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I want to think about this in $\mathbb{R}^1$ and $\mathbb{R}^2$.

In $\mathbb{R}^1$, isn't every interval a connected set? So if the intersection of two intervals is also an interval, that would be a connected set. If the intersection is a single point, that would also be connected. If the intersection is null, that is also a connected set.

Is it safe to say intersections of connected sets are always connected in $\mathbb{R}^1$?

For $\mathbb{R}^2$, intuitively, I think the answer is no. I'm not sure how to prove it though.

PS There's a similar question posted before the answer was for $\mathbb{R}^2$. Must the intersection of connected sets be connected? The answer "Consider the intersection of the line segment and the circle in $\varnothing$" I don't understand what they mean by a circle in $\varnothing$.

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mathinthecity
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    In the answer you linked to, $\varnothing$ isn't a mathematical symbol, it's a picture of a circle and a line segment. The circle O is connected; the line segment $/$ is connected; the intersection of the circle and the line segment is a two-point set, which is not connected. – bof Nov 07 '16 at 11:22

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In $\mathbb R$, you are correct. Every connected set is an interval, and the intersection of two intervals is a connected set.

Of course, to actually prove that, you have to prove that All connected sets in $\mathbb R$ are intervals, which is not too hard, but also not entirely trivial.

In $\mathbb R^2$, take $$A=\{(x,y)|x^2+y^2=1\}$$ and $$B=\{(x, 0)| x\in\mathbb R\}$$

Now:

  • What is the intersection of $A$ and $B$?
  • Is $A\cap B$ connected?
5xum
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  • Don't the two sets intersect at x=1? And therefore wouldn't the intersection be connected because it's a single point? – mathinthecity Nov 07 '16 at 11:20
  • @mathinthecity No. First of all, $x=1$ is not a point in $\mathbb R^2$. Points in $\mathbb R^2$ are of the form $(x,y)$, they have two numbers. So, $(4,4)$ is a point in $\mathbb R^2$, but $x=4$ is not. Second of all, there is another point at which they intersect. Try drawing $A$ and $B$. – 5xum Nov 07 '16 at 11:22
  • Oh, sorry. So they intersect at (-1,0) and (1,0) which is not a connected set. Thanks! – mathinthecity Nov 07 '16 at 11:25
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    @mathinthecity There you go. This is math, not the races. Try to slow your mind down when you start getting strange results. – 5xum Nov 07 '16 at 11:26
  • https://math.stackexchange.com/questions/239063/intervals-are-connected-and-the-only-connected-sets-in-mathbbr – Clemens Bartholdy Aug 03 '24 at 10:05