203: Topological Connected-ness of 2 Points Is Equivalence Relation

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that topological connected-ness of 2 points is equivalence relation

---

Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

Starting Context

• The reader knows a definition of topological connected-ness of 2 points.
• The reader knows a definition of equivalence relation.

---

Target Context

• The reader will have a description and a proof of the proposition that topological connected-ness of 2 points is an equivalence relation.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For any topological space, $T$, connected-ness of 2 points is an equivalence relation.

---

2: Proof

For any point, $p\in T$, $p$ and $p$ are connected, because $\{p\}$ is a topological subspace that is not the union of any disjoint open sets.

For any points, $p_1,p_2\in T$, that are connected, $p_2$ and $p_1$ are connected, because there is a connected topological subspace, $T_1$, that contains $p_1$ and $p_2$, which means that $T_1$ contains $p_2$ and $p_1$.

For any points, $p_1,p_2,p_3\in T$, such that $p_1$ and $p_2$ are connected and $p_2$ and $p_3$ are connected, there are a connected topological subspace, $T_1$, that contains $p_1$ and $p_2$ and a connected topological subspace, $T_2$, that contains $p_2$ and $p_3$. $T_1\cup T_2$ is a connected topological subspace that contains $p_1$ and $p_3$, as is proven as follows. Suppose that $T_1\cup T_2$ was not connected. $T_1\cup T_2=U_1\cup U_2$, $U_1\cap U_2=\mathrm{\varnothing }$ where $U_i$ would be a non-empty open set on $T_1\cup T_2$. $U_i={U_i}^{\prime }\cap (T_1\cup T_2)$ where ${U_i}^{\prime }$ would be open on $T$, by the definition of subspace topology. As ${U_1}^{\prime }\cup {U_2}^{\prime }$ would cover $T_1\cup T_2$, $T_i=T_i\cap ({U_1}^{\prime }\cup {U_2}^{\prime })=(T_i\cap {U_1}^{\prime })\cup (T_i\cap {U_2}^{\prime })$. ${U_1}^{\prime }$ and ${U_2}^{\prime }$ would not share any point on $T_i$, because otherwise, $U_1$ and $U_2$ would share the point. Each of both $T_1$ and $T_2$ would not be contained in a ${U_i}^{\prime }$, because if both $T_1$ and $T_2$ were contained in ${U_1}^{\prime }$ without loss of generality, $U_2={U_2}^{\prime }\cap (T_1\cup T_2)$ would be empty; if $T_1$ and $T_2$ were contained in ${U_1}^{\prime }$ and ${U_2}^{\prime }$ respectively without loss of generality, ${U_1}^{\prime }$ and ${U_2}^{\prime }$ would share $p_2$, then $U_1$ and $U_2$ would share $p_2$. So, at least 1 of $T_1$ and $T_2$ would be separated into ${U_1}^{\prime }$ and ${U_2}^{\prime }$, which would mean that at least 1 of $T_1$ and $T_2$ would not be connected, a contradiction.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>