1
$\begingroup$

Consider the set $A\subset \mathbb{R}^2$ of all rational points in the unit ball, that is $$A = \{(x,y)\in \mathbb{Q}^2: x^2+y^2<1\}.$$ It's easily seen that $A$ is not path-connected (since every continuous curve must pass through a point with at least one irrational coordinate, by the intermediate value theorem), so $A$ is not connected as well. Now consider the definition of connected subset:

A subset S of a topological space $X$ is connected if and only if there are no open sets $U$ and $V$ in $X$ such that $$S\subset U\cup V, S\cap U \neq \varnothing, S\cap V\neq \varnothing \text{ and } S\cap U\cap V = \varnothing.$$

Since $A$ is disconnected, there must be some open sets $U$ and $V$ in $\mathbb{R}^2$ such that $$A\subset U\cup V, A\cap U = \varnothing, A\cap V = \varnothing \text{ and } A\cap U \cap V = \varnothing.$$

Can we write specifically what $U$ and $V$ are?

Thank you very much.

$\endgroup$
2
  • 1
    $\begingroup$ Note that a not path-connected space does not have to be also not connected. See here: en.wikipedia.org/wiki/Connected_space#Path_connectedness. It may work in your example, but your reasoning is flawed for the general case. $\endgroup$
    – M. Winter
    Feb 23 '17 at 14:42
  • $\begingroup$ Thank you very much, I forgot the condition for $A$ to be open. $\endgroup$ Feb 23 '17 at 14:52
4
$\begingroup$

There are many examples of such a $U$ and $V$. One working example is $$ U = \{(x,y) \in \Bbb Q^2 : x^2 + y^2 < \frac 1{\sqrt{2}} \}\\ V = \{(x,y) \in \Bbb Q^2 : \frac 1{\sqrt{2}} < x^2 + y^2 < 1\} $$ The only important thing about $1/\sqrt{2}$ here is that it's irrational and less than $1$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.