Must a neighbourhood of a cut point $x$ of a topological space $X$ contain points from all path-components of $X\setminus\{x\}$?

Question

Must a neighbourhood $N$ of a cut point $x$ of a (path-connected) topological space $X$ contain points from all path-components of $X\setminus\{x\}$?

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The answer appears to be, clearly, yes. However, I can't think of an argument to justify this other than simply state that it follows immediately from the definition of a cut point. Is this the case, however? Or there's an obvious proof/counterexample I'm missing?

Answer 1

Let's try this example. But note that this example is not $T_1$! It might not be fair to be talking about a cut point if the space is not $T_1$.

When $X$ is $T_1$ then the proof in here shows that the statement is true.

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Take the space $X=\{2,3,4\}$ with the topology $T=\{\emptyset, \{3\}, \{2,3\}, \{3,4\}, X\}$.

With this topology $X$ is connected, since all non-empty open sets meet at $3$. The point $x=3$ is a cut point, since $Y=X\setminus \{3\}=\{2,4\}$ is disconnected by the (relative to $Y$) open sets $\{2,3\}\cap Y$ and $\{3,4\}\cap Y$.

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We can connect $2$ to $3$ by the path $p(t)=2$ for $t\in [0,1/2]$ and $p(t)=3$ for $t\in(1/2,1]$. This is continuous since $p^{-1}(\{3\})=(1/2,1]$ is open and $p^{-1}(\{2,3\})=[0,1]$ is also open.

We can connect $3$ and $4$ by $p(t)=4$ for $t\in[0,1/2]$ and $p(t)=3$ for $t\in(1/2,1]$. As above, $p(\{3\})=(1/2,1]$ is open and $p(\{3,4\})=[0,1]$ is also open.

We can connect $2$ to $4$ by $p(t)=2$ for $t\in[0,1/3]$, $p(t)=3$ for $t\in(1/3,2/3)$, and $p(t)=4$ for $t\in(2/3,1]$. This is continuous since $p^{-1}(\{2,3\})=[0,2/3)$ is open, $p^{-1}(\{3\})=(1/2,2/3)$ is open and $p^{-1}(\{3,4\})=(2/3,4]$ is also open.

So, $X$ is path connected.

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Now, the neighborhood $\{3\}$ of $3$ doesn't intersect $Y=X\setminus\{3\}$, let alone its path components.

Answer 2

Assume that $X$ is path-connected and $T_1$ and that $X \backslash \{x\}$ is not path-connected.

Let $U$ be a path component of $X \backslash \{x\}$ and let $u\in U$ be any point. Then there exists a point $v \in X\backslash\{x\}$ which is not in $U$.

Since $X$ is path-connected, there is a path (i.e. continuous map) $p:[0, 1] \rightarrow X$ such that $p(0) = u$ and $p(1) = v$.

But the path $p$ must pass the cut point $x$ - otherwise it would be a path in $X \backslash \{x\}$.

Therefore the inverse image $p^{-1}(x)$ is a non-empty closed subset of $(0, 1)$ (here we use the assumption that $X$ is $T_1$).

Let $t$ be the minimum element of $p^{-1}(x)$ (which exists by compactness). Since $N$ is a neighborhood of $x$ and $p$ is continuous, the inverse image $p^{-1}(N)$ contains an open subset of $(0, 1)$ containing $t$.

This means that there exists $0 < s < t$ such that $p(s)$ is in $N$. It only remains to show that $p(s)$ is in $U$.

Consider the sub-path $p\vert_{[0, s]}$. It connects the two points $u$ and $p(s)$. Moreover, this sub-path does not pass the point $x$, by minimality of $t$. Therefore $p(s)$ lies in the same path connected component as $u$, which is $U$.