A topological space $(X, \tau)$ is connected if $X$ is not the union of two nonempty, open, disjoint sets. A subset $Y \subseteq X$ is connected if it is connected in the subspace topology.

In detail, the latter means that there exist no (open) sets $A$ and $B$ in the original topology $\tau$ such that $Y \cap A \neq \emptyset$, $Y \cap B \neq \emptyset$, $Y \subseteq A \cup B$, $Y \cap A \cap B = \emptyset$.

Fine so far, but I am reading a text that defines a subset $Y$ of $\mathbb{R}^n$ with its usual topology to be connected by keeping the first three requirements, but making the final one more stringent: $A \cap B = \emptyset$.

Is that definition really equivalent (i) in $\mathbb{R}^n$ and (ii) in general topological spaces?

  • $\begingroup$ You can assume that $A, B \subseteq Y$ without loss of generality by the definition of the subspace topology. $\endgroup$ Sep 15 '14 at 8:38
  • 1
    $\begingroup$ @QiaochuYuan: I don't think so. What about sets with empty interior? $\endgroup$
    – Crostul
    Sep 15 '14 at 8:55
  • $\begingroup$ Ah, tricky. I guess I need $Y$ to be open for the argument I had in mind to work. $\endgroup$ Sep 15 '14 at 20:59

When dealing with the subspace topology, one really shouldn't bother what happens outside the subspace in consideration. Nevertheless, assume that $Y$ is a subset of $\mathbb R^n$ such that there exist open sets $A,B$ such that $Y\cap A$ and $Y\cap B$ are disjoint and nonempty, but necessarily $A\cap B\ne \emptyset$ for such $A,B$ (clearly, the intersection is outside of $Y$). We shall see that this cannot happen. All we use to show this is that the topology of $\mathbb R^n$ comes from a metric:

Since $A$ is open, for every point $x\in Y\cap A$, there exists $r>0$ such thet the $r$-ball $B(x,r)$ around $x$ is $\subseteq A$ and hence disjoint from $B\cap Y$. Pick such $r=r(x)$ for each $x\in A\cap Y$ and similarly for each $x\in B\cap Y$ such that $B(x,r)$ is disjoint from $A\cap Y$. Then $A':=\bigcup_{x\in A\cap Y}B(x,\frac12r(x))$ and $B':=\bigcup_{x\in B\cap Y}B(x,\frac12r(x))$ are open subsets of $\mathbb R^n$, are disjoint, $A'\cap Y=A\cap Y$, and $B'\cap Y=B\cap Y$.

In a general topological space, the situation may differ. Let $X=\{a,b,c\}$ where a subset is open iff it is empty or contains $c$. Then the subspace topology of $Y=\{a,b\}$ is discrete, hence $Y$ is not connected, but any open $A,B\subset X$ witnessing this of course intersect in $c$.


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