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I think the statement is true that two connected subspaces of a connected topological space is connected, and there are two different situations here to be discussed.

First of all, when the intersection is empty set, do we consider it as connected or not?

For the other situation, if the intersection is not empty, how could I get the conclusion?

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  • $\begingroup$ I suppose you mean the union of two connected subspaces? $\endgroup$
    – Xuqiang QIN
    Commented Dec 12, 2015 at 23:25
  • $\begingroup$ What if the space is the real line, and the two connected subspaces are the intervals $(0,1)$ and $(2,3)$? $\endgroup$
    – bof
    Commented Dec 12, 2015 at 23:26
  • $\begingroup$ @XuqiangQin Sorry for the confusion here, I missed the intersection on the question part. $\endgroup$
    – Luker
    Commented Dec 12, 2015 at 23:27
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    $\begingroup$ @Luker The empty set is connected. Otherwise by deifnition it would be the union of two nonempty disjoint sets, which is impossible. $\endgroup$
    – Forever Mozart
    Commented Dec 13, 2015 at 6:59
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    $\begingroup$ A simple example: $\between$ is the intersection of $($ and $)$. $\endgroup$
    – Brian M. Scott
    Commented Dec 13, 2015 at 19:05

2 Answers 2

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The empty set is connected (trivially), because we cannot write it as a union of non-empty (!) disjoint open sets..

But the intersection of two connected sets need not be connected at all. Consider $C = \{(x,y): x^2 + y^2 = 1\}$, which is the unit circle (connected) and $D = \{(x,y): (x-1)^2 + y^2 = 1 \}$, the circle of radius 1 around $(1,0)$, also connected. Their intersection is $\{(\frac{1}{2},\frac{1}{2}\sqrt{3}),(\frac{1}{2},-\frac{1}{2}\sqrt{3})\}$, which is a two point set in the plane, hence disconnected.

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    $\begingroup$ Is it possible to construct an example in $\Bbb R$ ? $\endgroup$
    – Empty
    Commented Nov 2, 2017 at 6:44
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    $\begingroup$ @S717717 no, in the reals connected is the same as order convex (intervals or segments ) and these are closed under intersections. $\endgroup$
    – Henno Brandsma
    Commented Nov 2, 2017 at 6:50
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Consider the intersection of the line segment and the circle in $\mathbb R^2$.

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