# Show that $\{a\}$ is connected set.

Let $(X,d)$ a metric space and $a\in X$. Show that $\{a\}$ is connected set.

My approach: Let $a\in X$, and suppose that $\{a\}$ is not connected set,i.e., there exist open set $A,B\subset X$ non-empty, such that $\{a\}\subseteq A\cup B$ and $A\cap B=\emptyset$. Thus implies that, if $\{a\}\in A$ then $B=\emptyset$; and if $\{a\}\in B$ then $A=\emptyset$. So this separation of the space cannot exist. Then $\{a\}$ is a connected set. This is correct? Thanks!!!

• You could also state that there cannot exist two disjoint non empty sets $A,B$ such that $A\cup B = \{a\}$ hence $\{a\}$ is connected. Oct 11 '17 at 3:52

There is a thing wrong in your argument. If you are to prove or disprove the whole space $X$ is connected, you use the definition "there exists two disjoint and nonempty open sets in $X$ of which the union is the whole space". But to prove or disprove a subspace $S$ of $X$ is connected, you use an alternative definition "there exists two disjoint and nonempty open sets in $X$ of which the union is a superset of $S$".
More precisely, in the second line of your paragraph two, you should write $\{a\}\subseteq A\cap B$ instead of $\{a\}=A\cap B$.
To give an example demonstrating how using $S=A\cap B$ can be problematic, consider $\Bbb R$ with the Euclidean metric, and try to show that $\{0,1\}\subseteq\Bbb R$ is disconnected.
• The argument is not wrong, one just uses the subspace topology. In the last paragraph, the sets $\{0\}, \{1\}$ are open in the subspace topology. Oct 11 '17 at 3:54
• @copper.hat Read his argument carefully. He picked open sets $A,B\subset X$, i.e. he is not finding open sets in the subspace.