I am trying to prove that in a set $X$ with discrete topology, if $X$ is connected, then it contains exactly one element.
Here is my proof:
Suppose that $X$ is connected and suppose as a contradiction that it has at least two elements. Then, consider the elements in X in the following way. Consider the subset $X_1$, which consists of just one element, and consider the subset $X_2=X_1^{c}$. Then, we have that $X_1\neq \emptyset$ and $X_2 \neq \emptyset$, we also have that $X_1 \cap X_2 \neq \emptyset$. Now, we show $X_1$ is open. So for any arbitrary $x \in X_1$, $\exists \delta=1$ such that $B(x,\delta)\subset X_1$, this is true since $X_1$ contains only one element, so $B(x,\delta=1)={x}\subset X_1$.
Now, we need to show $X_2$ is open. So, pick any $x \in X_2$, then let $\delta=2$ and for such $x$, consider the ball $B(x,\delta)$. Since the set $X_2$ has the metric topology, then all the elements different from $x$ will be distance 1 away from $x$ in $X_2$, so $B(x,\delta=2)\subset X_2$ for any $x \in X_2$ and so $X_2$ will be open.
Since $X_1$ and $X_2$ are open, then by definition $X$ should be disconnected, which contradicts our assumption that it is connected and so it must be that X contains only 1 element.
Is this proof right? What am I missing?
Best Answer
Your proof is correct if you assume that the space is metric, but that is in fact just an unnecessary complication. A more general proof (but along exactly the same lines) is:
If $X$ contains at least two points, then choose any point $*\in X$ and take $$A = \{*\},\qquad B=X\backslash\{*\}.$$ As we are working in the discrete topology, every subset of $X$ is open, so $(A,B)$ form an open partition of $X$, showing that $X$ is not connected.
Conversely, if $X$ consists of a single point, then it is obviously connected.