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

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

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?

• Can you please add some details to your question? What, precisely, is your definition of a cut point? What kinds of topological spaces are you considering (must they be path-connected)? Commented Nov 28, 2019 at 0:21
• @XanderHenderson Yes, path-connected. Here the definition I'm using: en.wikipedia.org/wiki/Cut-point Commented Nov 28, 2019 at 0:26
• Please edit your question to include this context. Commented Nov 28, 2019 at 0:27
• @XanderHenderson Could you remove the closing request on the thread? It's not justified. Commented Nov 28, 2019 at 0:30
• The close votes to this post is really disappointing. I do not see anything missing.
– user9464
Commented Nov 28, 2019 at 12:00

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 $$U$$.

• The problem in your argument (at least the first) is in the third paragraph. Connectedness of $X$ doesn't imply existence of a path. Commented Nov 28, 2019 at 0:45
• @conditionalMethod I somehow assumed that $X$ is path-connected, and it was added to the question. However the removal of the metric-spaces tag might invalidate my argument that $\{x\}$ is a closed set. Commented Nov 28, 2019 at 0:48
• Path connectedness of X is implied, yes. Commented Nov 28, 2019 at 0:50
• Well, if they change the hypotheses the question changes. Commented Nov 28, 2019 at 0:50
• @Stephen You mean, given, right? Commented Nov 28, 2019 at 0:51

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.

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 $$\{3,4\}\cap Y$$.

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.

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

• Right. That's why I was worrying about the condition $\{x\}$ being closed. P.S. after connecting $2, 3$ and $3, 4$, you don't need to connect $2,4$ manually, since path connectedness is an equivalence relation. Commented Nov 28, 2019 at 1:53
• Thanks. This is a very revealing example. It shows how one goes about creating paths and proving their continuity in rather contrived situations. Commented Nov 28, 2019 at 1:57
• @WhatsUp At least in Wikipedia the definition of cut point does include the assumption that the space is $T_1$. Commented Nov 28, 2019 at 1:57
• @WhatsUp But then in the same page, the first property in here doesn't assume it. Anyway, the exact form of a definition is not very important but to understand how the different properties interact when one has them or don't have them. Commented Nov 28, 2019 at 2:02
• @Stephen Well, the fact that one can get the non-intersection part very cheaply if the cut point is open makes you try to see if all other properties can also be obtained. Similar to when $X$ was only connected, the fact that path-connected components can be different from connected components makes you think that in that case something can be done to make it not work. Commented Nov 28, 2019 at 2:32