Let $X$ be a set and assume $p\in X$. Prove that every subset of $X$ is connected in the particular point topology on $X$ and in the excluded point topology on $X$.
http://en.wikipedia.org/wiki/Particular_point_topology
http://en.wikipedia.org/wiki/Excluded_point_topology
For the particular point topology I've come this far:
Let $A$ be a subset of $X$. Suppose $p\in A$. Any open set can be written as $U=V \cap A$ with V being open in X. By definition of the particular point topology $p\in V\Rightarrow p\in U$, hence any open set in $A$ contains $p$, and A can therefore not be represented as a disjoint union of non-empty open sets.
But what if $p\notin A$ ?
For the excluded point I have this much:
Let $A$ be a subset of $X$ with $p\in A$. Then since any open set $U$ in $A$ can be written as $U=V\cap A$ where $V$ is open in $X$. By the defn. of the excluded point topology any open set in $X$ does not contain $p$, implying that $U$ does not contain $p$. Consequently $A$ is connected as no open subsets in $A$ contains $p$.
Again I'm stucked when $p\notin A$?
Best Answer
I don't think these claims are actually true.
Let $X$ have the particular point topology. Suppose $A$ is a subset of $X$ with $p \notin A$. Then any subset $U$ of $A$ can be written as $$U = (U \cup \{p\}) \cap A$$ and is therefore open. It follows that the subspace topology on $A$ is the discrete topology. Hence $A$ is disconnected.
Now let $X$ have the excluded point topology. Suppose $A$ is a subset of $X$ with $p \notin A$. Then any subset $U$ of $A$ is open in $A$ since it is already open in $X$. Again, it follows that the subspace topology is discrete.
Your proofs for when $p \in A$ are mostly correct. But in the proof for the excluded point topology, you should probably say that the only open set in $X$ which contains $p$ is $X$, and so the only open set in $A$ which contains $p$ is $A$.