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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 $p\notin A$?

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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 $A$.

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Thank you for the answer. It seems that you are right that the claims are false. –  Peter Smoke Nov 8 '13 at 15:57
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