Looks to me that it should be true, although sets like $A = \{\frac{1}{n}, n \in \mathbb{N}\}$ give me doubts
edit: I know it isn't a counter example but the fact that there exist sets such that their set of limit points is a point gives me doubts
Asked
Active
Viewed 745 times
0
-
1Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Aug 06 '19 at 08:13
-
1What do you mean A={1\n, n from N}? – Bach Aug 06 '19 at 08:15
-
The set ${ \frac1n\colon n\in\Bbb N}$ is not connected, and its set of limit points ${0}$ is connected; so this is far from a counterexample. – Greg Martin Aug 06 '19 at 08:21
1 Answers
4
Let $A$ be a connected set in a metric space. Then $A$ has no isolated points: if $x$ is an isolated point then $\{x\}$ is open and closed in $A$ contradicting connectedness. Hence the set of limit points of $A$ is nothing but the closure of $A$. The closure of a connected set is always connected.
Kavi Rama Murthy
- 311,013