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I do have problems with connected sets so I got the following exercise:

$X:=\{x\in\mathbb{R}^n: \|x\|=\pi\}\cup\{0\}\subset\mathbb{R}^n$. Why is $X$ not connected?

My attempt: I have to find disjoint open sets $U,V\ne\emptyset$ such that $U\cup V=X$ .

Let $U=\{x\in\mathbb{R}^n: \|x\|=\pi\}$ and $V=\{0\}$. Then $V$ is relative open since $$V=\{x\in\mathbb{R}^n:\|x\|<1\}\cap X$$ and $\{x\in\mathbb{R}^n:\|x\|<1\}$ is open in $\mathbb{R}^n$.

Is this right? and why is $U$ open?

Zev Chonoles
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5 Answers5

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Yup, that's right! And similarly, you can see that $U$ is open because $U=W\cap X$ where $$W=\{x\in\mathbb{R}^n:\tfrac{\pi}{2}<\|x\|<\tfrac{3\pi}{2}\}$$ is an open set of $\mathbb{R}^n$. One way of seeing that $W$ is open is noting that the function $\|\cdot\|:\mathbb{R}^n\to\mathbb{R}$ is continuous, and $W$ is the preimage of the open set $(\frac{\pi}{2},\frac{3\pi}{2})$.

Zev Chonoles
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  • I have a stupid question: this relies on $X$ being open, right? – Lucas Jun 11 '13 at 23:30
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    @Lucas: Sorry, I should have been clearer: $U$ is relatively open in $X$, but $U$ is not open as a subset of $\mathbb{R}^n$. It is relatively open $X$, by definition, because it is the intersection of an open set of $\mathbb{R}^n$ (namely $W$) with $X$. – Zev Chonoles Jun 12 '13 at 00:52
  • OK, I'm still a little confused (my problem entirely). Is the general idea with these problems to say that there is an set that is a superset of the set you are testing, and that set is open, so as long as your can say these supersets are disconnected, you can say the originals are? – Lucas Jun 12 '13 at 02:09
  • @Lucas: It may help you to look at the Wikipedia page on the subspace topology. Remember, "open" is not an inherent property of a set, it depends on the ambient space we're working in (which will often be left implicit). Properly speaking, we need a set $Y$ and a topology $T$ on $Y$; only then we can make sense of the statement "$Z$ is an open subset of $Y$". So, when I say "$U$ is open", what I mean is "$U$ is an open subset of $X$, where $X$ has been given the subspace topology from $\mathbb{R}^n$". – Zev Chonoles Jun 12 '13 at 02:23
  • To demonstrate that the set $X$ in this problem is disconnected, we need to demonstrate the existence of two disjoint, non-empty open subsets of $X$ whose union is $X$. Again, "open subset of $X$" here means "subset of $X$ that is open in the subspace topology $X$ gets from $\mathbb{R}^n$". – Zev Chonoles Jun 12 '13 at 02:26
  • This will help me: is $X$ relatively open in $X$? – Lucas Jun 12 '13 at 03:40
  • @Lucas: Yes, that's true. If $Y$ is our beginning topological space, then $Y$ is open as a subset of $Y$ by the definition of a topology. Therefore, for any given subset $X\subseteq Y$ whatsoever, $X$ is the intersection of an open set of $Y$ with $X$: namely, $X=X\cap Y$. – Zev Chonoles Jun 12 '13 at 03:46
  • Sweet, cool, I get it. Thanks very much. – Lucas Jun 12 '13 at 03:49
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Yes, that is right. And $U=\{x\in\mathbb{R}^n\colon\|x\|>0\}\cap X$.

Harald Hanche-Olsen
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A more fancy approach: it will suffice to say that the singleton $\{0\}$ is a clopen (=closed and open) set. It is closed, because one point sets are always closed. It is open, because it is $X \cap \{ x \in \mathbb{R}^n \ : \ || x || < 1 \}$

Jakub Konieczny
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4

Alternative: the norm $\|\cdot \|$ is 1-Lipschitz by the triangular inequality, whence continuous. The image of your set under this continuous map is $\{0\}\cup\{\pi\}$, which is disconnected. So your set can't be connected, since the continuous image of a connected space is connected.

Julien
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Any open set O in $\mathbb R^n$ intersected with $X$ yields an open set $O\cap X$ in X.

Proof: by definition, a set $S\subseteq X$ is open in $X$, iff for each $x \in S$ an $\epsilon>0$ exists such that the ball $B(x,X,\epsilon)= \{ y \in X : ||x-y||< \epsilon\} \subseteq S$. Now if $S=O\cap X$ for an open set $O \subseteq \mathbb R^n$, then for each $x \in S$, there exists an $\epsilon$ such that $B(x,\mathbb R^n,\epsilon) \subseteq O$. This implies that $B(x,X,\epsilon)=B(x,\mathbb R^n,\epsilon)\cap X \subseteq O\cap X=S$.

So $V$ is open. Similarly, $U= \{x\in \mathbb R^n: 3<||x||<4\}\cap X$ is also open in $X$. As $X$ is the disjoint union of two open sets, it is not connected.

vinnief
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