# limit points and closure for connected sets

The below theorem shows that adding limit points^{} to a connected
set preserves connectedness.

###### Theorem 1.

Suppose $A$ is a connected set in a topological space^{}.
If $A\mathrm{\subseteq}B\mathrm{\subseteq}\overline{A}$, then $B$ is connected.
In particular, $\overline{A}$ is connected.

Thus, one way to prove that a space $X$ is connected is to find a dense subspace in $X$ which is connected.

Two touching closed balls^{} in ${\mathbb{R}}^{2}$ shows that this theorem does not hold
for the interior. Along the same lines, taking the closure^{} does not
preserve separatedness.

###### Proof.

Let $X$ be the ambient topological space.
By assumption^{}, if $U,V\subseteq A$ are open and $U\cup V=A$, then
$U\cap V\ne \mathrm{\varnothing}$.
To prove that $B$ is connected, let $U,V$ be open sets in
$B$ such that $U\cup V=B$ and for a
contradition, suppose that $U\cap V=\mathrm{\varnothing}$.
Then there are open sets $R,S\subseteq X$ such that

$$U=R\cap B,V=S\cap B.$$ |

It follows that $(R\cup S)\cap B=B$ and $(R\cap S)\cap B=\mathrm{\varnothing}$. Next, let $\stackrel{~}{U},\stackrel{~}{V}$ be open sets in $A$ defined as

$$\stackrel{~}{U}=R\cap A,\stackrel{~}{V}=S\cap A.$$ |

Now

$$A=B\cap A\subseteq (R\cup S)\cap A\subseteq A$$ |

and as $(R\cup S)\cap A=\stackrel{~}{U}\cup \stackrel{~}{V}$, it follows that $\mathrm{\varnothing}\ne \stackrel{~}{U}\cap \stackrel{~}{V}=(R\cap S)\cap A$. Then, by the properties of the closure operator,

$$\mathrm{\varnothing}\ne \overline{(R\cap S)\cap A}\supseteq (R\cap S)\cap \overline{A}\supseteq (R\cap S)\cap B=\mathrm{\varnothing}.$$ |

∎

Title | limit points and closure for connected sets |
---|---|

Canonical name | LimitPointsAndClosureForConnectedSets |

Date of creation | 2013-03-22 15:17:56 |

Last modified on | 2013-03-22 15:17:56 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 7 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 54D05 |