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I don't know if this answers your question precisely but here's an interesting example. First, let's start with the following:

Knaster-Kuratowski Example

Let $K$ be the Knaster-Kuratowski fan, also called the Cantor's teepee. The space $K$ is defined as follows: let $C$ be the Cantor Set. Let $Q$ be the set of endpoints of the deleted middle-third intervals. Then $Q \subset C$. Let $P=C \backslash Q$. We also let $p=(1/2,1/2) \in \mathbb R^2$. Now $\forall x \in C$, we let $L_x$ be the line joining $p$ and $x$. Now our space $K$ is the union over all $x \in C$ of the lines $L_x=\ {(x,y):y\in \mathbb Q \ }$ if $x\in Q$ and $L_x=\ {(x,y):y\in Irr \ }$ if $x \in P$. The space $K$ is connected but $K\backslash {p}$ is totally disconnected. Also $K$ is punctiform, that is it contains no compact connected T2 subspace.

Answer, a related property.

Let $f$ be a continuous function from $K$ to $K$. Let $U$ be some closed connected open set about a point $x$, $x \neq p$, then there exists a closed connected open set $V$ about $p$, which can be written down as the set of all $(x,y)$ with $1/2 - \epsilon < y < 1/2$ such that $f(V) \subset U$ since $f$ is continuous. Then $V$ is necessarily homeomorphic to the space $K$

The author of this answer: Carlo Von Schnitzel

I don't know if this answers your question precisely but here's an interesting example. First, let's start with the following:

The Knaster-Kuratowski Fan

Let $K$ be the Knaster-Kuratowski fan, also called Cantor's teepee. The space $K$ is defined as follows. Let $C$ be the Cantor set. Let $Q \subset C$ be the set of endpoints of the deleted middle-third intervals. Let $P=C \setminus Q$. We also let $p=(1/2,1/2) \in \mathbb R^2$. Now for each $x \in C$, we let $L_x$ be the line joining $p$ and $x$. Now our space $K$ is the union over all $x \in C$ of the sets

• $\{ (x,y) \in L_x : y\in \mathbb Q \}$, if $x\in Q$, and
• $\{ (x,y) \in L_x : y\notin \mathbb Q \}$, if $x \in P$.

The space $K$ is connected but $K\setminus \{p\}$ is totally disconnected. Also $K$ is punctiform, that is it contains no compact connected T₂ subspace.

A related property of $K$

Let $f$ be a continuous function from $K$ to $K$. Let $U$ be some closed connected open set about a point $x \neq p$. Then there exists a closed connected open set $V$ about $p$, which can be written down as the set of all $(x,y)$ with $1/2 - \epsilon < y < 1/2$ such that $f(V) \subset U$ since $f$ is continuous. Then $V$ is necessarily homeomorphic to the space $K$.

I don't know if this answers your question precisely but here's an interesting example. First, let's start with the following:

Knaster-Kuratowski Example

Let $K$ be the Knaster-Kuratowski fan, also called the Cantor's teepee. The space $K$ is defined as follows: let $C$ be the Cantor Set. Let $Q$ be the set of endpoints of the deleted middle-third intervals. Then $Q \subset C$. Let $P=C \backslash Q$. We also let $p=(1/2,1/2) \in \mathbb R^2$. Now $\forall x \in C$, we let $L_x$ be the line joining $p$ and $x$. Now our space $K$ is the union over all $x \in C$ of the lines $L_x=\ {(x,y):y\in \mathbb Q \ }$ if $x\in Q$ and $L_x=\ {(x,y):y\in Irr \ }$ if $x \in P$. The space $K$ is connected but $K\backslash {p}$ is totally disconnected. Also $K$ is punctiform, that is it contains no compact connected T2 subspace.

Answer, a related property.

Let $f$ be a continuous function from $K$ to $K$. Let $U$ be some closed connected open set about a point $x$, $x \neq p$, then there exists a closed connected open set $V$ about $p$, which can be written down as the set of all $(x,y)$ with $1/2 - \epsilon < y < 1/2$ such that $f(V) \subset U$ since $f$ is continuous. Then $V$ is necessarily homeomorphic to the space $K$

The author of this answer: @CarloVonSchnitze

I don't know if this answers your question precisely but here's an interesting example. First, let's start with the following:

Knaster-Kuratowski Example

Let $K$ be the Knaster-Kuratowski fan, also called the Cantor's teepee. The space $K$ is defined as follows: let $C$ be the Cantor Set. Let $Q$ be the set of endpoints of the deleted middle-third intervals. Then $Q \subset C$. Let $P=C \backslash Q$. We also let $p=(1/2,1/2) \in \mathbb R^2$. Now $\forall x \in C$, we let $L_x$ be the line joining $p$ and $x$. Now our space $K$ is the union over all $x \in C$ of the lines $L_x=\ {(x,y):y\in \mathbb Q \ }$ if $x\in Q$ and $L_x=\ {(x,y):y\in Irr \ }$ if $x \in P$. The space $K$ is connected but $K\backslash {p}$ is totally disconnected. Also $K$ is punctiform, that is it contains no compact connected T2 subspace.

Answer, a related property.

Let $f$ be a continuous function from $K$ to $K$. Let $U$ be some closed connected open set about a point $x$, $x \neq p$, then there exists a closed connected open set $V$ about $p$, which can be written down as the set of all $(x,y)$ with $1/2 - \epsilon < y < 1/2$ such that $f(V) \subset U$ since $f$ is continuous. Then $V$ is necessarily homeomorphic to the space $K$

The author of this answer: Carlo Von Schnitzel

I don't know if this answers your question precisely but here's an interesting example.

Let $K$ be the Knaster-Kuratowski fan, also called the Cantor's teepee. The space $K$ is defined as follows: let $C$ be the Cantor Set. Let $Q$ be the set of endpoints of the deleted middle-third intervals. Then $Q \subset C$. Let $P=C \backslash Q$. We also let $p=(1/2,1/2) \in \mathbb R^2$. Now $\forall x \in C$, we let $L_x$ be the line joining $p$ and $x$. Now our space $K$ is the union over all $x \in C$ of the lines $L_x=\ {(x,y):y\in \mathbb Q \ }$ if $x\in Q$ and $L_x=\ {(x,y):y\in Irr \ }$ if $x \in P$. The space $K$ is connected but $K\backslash {p}$ is totally disconnected. Also $K$ is punctiform, that is it contains no compact connected T2 subspace. Let $f$ be a continuous function from $K$ to $K$. Let $U$ be some closed connected open set about a point $x$, $x \neq p$, then there exists a closed connected open set $V$ about $p$, which can be written down as the set of all $(x,y)$ with $1/2 - \epsilon < y < 1/2$ such that $f(V) \subset U$ since $f$ is continuous. Then $V$ is necessarily homeomorphic to the space $K$

I don't know if this answers your question precisely but here's an interesting example. First, let's start with the following:

Knaster-Kuratowski Example

Let $K$ be the Knaster-Kuratowski fan, also called the Cantor's teepee. The space $K$ is defined as follows: let $C$ be the Cantor Set. Let $Q$ be the set of endpoints of the deleted middle-third intervals. Then $Q \subset C$. Let $P=C \backslash Q$. We also let $p=(1/2,1/2) \in \mathbb R^2$. Now $\forall x \in C$, we let $L_x$ be the line joining $p$ and $x$. Now our space $K$ is the union over all $x \in C$ of the lines $L_x=\ {(x,y):y\in \mathbb Q \ }$ if $x\in Q$ and $L_x=\ {(x,y):y\in Irr \ }$ if $x \in P$. The space $K$ is connected but $K\backslash {p}$ is totally disconnected. Also $K$ is punctiform, that is it contains no compact connected T2 subspace.

Answer, a related property.

Let $f$ be a continuous function from $K$ to $K$. Let $U$ be some closed connected open set about a point $x$, $x \neq p$, then there exists a closed connected open set $V$ about $p$, which can be written down as the set of all $(x,y)$ with $1/2 - \epsilon < y < 1/2$ such that $f(V) \subset U$ since $f$ is continuous. Then $V$ is necessarily homeomorphic to the space $K$

The author of this answer: @CarloVonSchnitze