2
$\begingroup$

Is there a connected $T_2$ space $(X,\tau)$ with more than one point, such that the singletons and $X$ are the only connected subspaces of $X$?

$\endgroup$
6
$\begingroup$

There is no such space. For if $x$ is any point and $X\setminus \{x\}=U|V$ then $\{x\}\cup U$ and $\{x\}\cup V$ are connected sets, each with more than one point and different from $X$.

The closest thing you can get is a connected set whose connected subsets are cofinite. The axiom CH implies there is a countable connected Hausdorff space with this property here that is not too complicated (relative to the completely regular example). I don't think a Hausdorff example has ever been constructed in ZFC.

$\endgroup$
  • 1
    $\begingroup$ What is $X\setminus \{x\}=U|V$? $\endgroup$ – Mad Hatter Nov 14 '17 at 7:33
  • $\begingroup$ Assuming $X\setminus \{x\}$ is not connected (otherwise the space fails to have the desired property), it is equal to the union of two disjoint nonempty open sets $U$ and $V$. $\endgroup$ – D.S. Lipham Nov 14 '17 at 7:35
  • 2
    $\begingroup$ A standard notation for disjoint unions is $U\sqcup V$ ($\backslash$sqcup) $\endgroup$ – YCor Nov 14 '17 at 10:59
  • 7
    $\begingroup$ @erz If $U\cup\{x\}$ is not connected, write it as $W\sqcup Y$ with $x\in Y$, $W,Y$ both nonempty and closed in $U\cup\{x\}$. Then $X=W\sqcup (Y\cup V)$ and both are closed and nonempty. $\endgroup$ – YCor Nov 14 '17 at 11:01
  • 1
    $\begingroup$ Incidentally, this shows that every infinite connected Hausdorff space admits an infinite connected proper subset which is either open (as the complement of a singleton) or closed. Iterating, we obtain a strictly decreasing sequence of connected subsets, each of which has finite complement in its closure. $\endgroup$ – YCor Nov 14 '17 at 13:28

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.