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If $Y$ is a subspace of $X$, and $C$ is a connected component of $Y$, then C need not be a connected component of $X$ (take for instance two disjoint open discs in $\mathbb{R}^2$).

But I read that, under the same hypothesis, $C$ need not even be connected in $X$. Could you please provide me with an example, or point me towards one?

Thank you.

SOURCE http://www.filedropper.com/manifolds2 Page 129, paragraph following formula (A.7.16).

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What do you mean by 'connected in $X$'? Connected viewed as a subspace of $X$? –  LostInMath Oct 7 '11 at 18:45
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connectedness is like compactness an intrinsic property: it does not matter how a subspace sits in a space. –  Henno Brandsma Oct 7 '11 at 19:56

3 Answers 3

Isn't it just false? The image of a connected subspace by the injection $Y\longrightarrow X$ is connected...

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Are you sure you read correctly? Suppose $C$ is not connected in $X$. By definition this means that there exists two open sets $U, V\subset X$ such that $U\cap C \neq \emptyset$, $V\cap C\neq \emptyset$, $U\cap V = \emptyset$, and $C\subset U\cup V$. But then by definition of subspace topology, $U' = U\cap Y$ is open in $Y$, and $V' = V\cap Y$ is open in $Y$. And since $C\subset Y$, you have that $U' \cap V' = \emptyset$ and $C \subset U'\cup V'$, and $C \subset C\cap U \subset Y\cap U = U'$ etc. so $C$ cannot be connected in $Y$.

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I'm probably wrong, but, how do you know that $U'$ and $V'$ are not empty? –  Weltschmerz Oct 7 '11 at 16:05
    
@Weltschmerz: $C\subset Y$, and $C\cap U\neq \emptyset$ $\implies$ $C\cap U \subset Y\cap U = U' \neq \emptyset$. And mutatis mutandis for $V'$. –  Willie Wong Oct 7 '11 at 18:35
    
@Wille: right, thank you. I posted the source, if you're interested. –  Weltschmerz Oct 7 '11 at 18:54
    
You need not to have $U \cap V = \emptyset$. What you need is $U \cap V \cap C = \emptyset$. See my "wrong" reply. –  André Caldas Oct 10 '11 at 12:08

As @LostInMath pointed, the example below is wrong because in the definition of disconnected, the open sets do NOT need to be disjoint. The example below just shows that a set might be disconnected and yet there are no disjoint open sets separating them.


I guess what you want is this:

Let $C \subset Y \subset X$. Even if $C$ is connected in $X$, it doesn't mean it is connected as a subset of $Y$. In fact, it is possible that $Y$ is connected as a subset of $X$ but disconnected as a topological space.

Let $X = \{a, b, c\}$ with the topology given by the sets $\emptyset, X, \{a,b\}, \{b,c\}$. And let $Y = \{a,c\}$. No $X$ subset is disconnected since every non-empty open set has $b$ as a common element. But $Y$ is disconnected. In fact, it is discrete.

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I think that 'no subset of $X$ is disconnected' and '$Y$ is disconnected' are contradicting each other, and the first is false while the second is true. –  LostInMath Oct 8 '11 at 21:58
    
@LostInMath: Sorry, I mixed this up with "completely normal". I will edit the post to say it is wrong. Who is the "lost in math" now? :-P –  André Caldas Oct 10 '11 at 12:02

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