850: For Topological Space, Point, and Neighborhood of Point, Neighborhood of Point on Neighborhood Is Neighborhood of Point on Base Space

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

description/proof of that for topological space, point, and neighborhood of point, neighborhood of point on neighborhood is neighborhood of point on base space

---

Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of neighborhood of point.
• The reader knows a definition of subspace topology of subset of topological space.

---

Target Context

• The reader will have a description and a proof of the proposition that for any topological space, any point, and any neighborhood of the point, any neighborhood of the point on the neighborhood is a neighborhood of the point on the base space.

---

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: Structured Description

Here is the rules of Structured Description.

Entities:
$T$: $\in \{\text{ the topological spaces }\}$
$p$: $\in T$
$N_p^{\prime }$: $\in \{\text{ the neighborhoods of }p\text{ on }T\}$
$N_p$: $\in \{\text{ the neighborhoods of }p\text{ on }{N}_p^{\prime }\}$
//

Statements:
$N_p$: $\in \{\text{ the neighborhoods of }p\text{ on }T\}$
//

---

2: Note

This proposition is less obvious than that any open neighborhood of the point on any open neighborhood is an open neighborhood of the point on the base space.

---

3: Proof

Whole Strategy: Step 1: take an open neighborhood of $p$ on $T$, $U_p^{\prime }\subseteq T$, such that $U_p^{\prime }\subseteq N_p^{\prime }$; Step 2: take an open neighborhood of $p$ on $N_p^{\prime }$, $U_p\subseteq N_p^{\prime }$, such that $U_p\subseteq N_p$ and an open neighborhood of $p$ on $T$, $U_p^{″}\subseteq T$, such that $U_p=U_p^{″}\cap N_p^{\prime }$; Step 3: see that $U_p^{\prime }\cap U_p^{″}\subseteq T$ is an open neighborhood of $p$ on $T$ such that $U_p^{\prime }\cap U_p^{″}\subseteq N_p$.

Step 1:

As $N_p^{\prime }$ is a neighborhood of $p$ on $T$, there is an open neighborhood of $p$ on $T$, $U_p^{\prime }\subseteq T$, such that $p\in U_p^{\prime }\subseteq N_p^{\prime }$.

Step 2:

As $N_p$ is a neighborhood of $p$ on $N_p^{\prime }$, there is an open neighborhood of $p$ on $N_p^{\prime }$, $U_p\subseteq N_p^{\prime }$, such that $U_p\subseteq N_p$.

According to the definition of subspace topology, there is an open neighborhood of $p$ on $T$, $U_p^{″}\subseteq T$, such that $U_p=U_p^{″}\cap N_p^{\prime }$.

Step 3:

Let us see that $U_p^{\prime }\cap U_p^{″}\subseteq T$ is an open neighborhood of $p$ on $T$ such that $U_p^{\prime }\cap U_p^{″}\subseteq N_p$.

$p\in U_p^{\prime }\cap U_p^{″}$.

$U_p^{\prime }\cap U_p^{″}$ is an open subset of $T$ as a finite intersection of open subsets.

So, $U_p^{\prime }\cap U_p^{″}$ is an open neighborhood of $p$ on $T$.

$U_p^{\prime }\cap U_p^{″}\subseteq N_p^{\prime }\cap U_p^{″}=U_p\subseteq N_p$.

So, $N_p$ satisfies the requirement to be an open neighborhood of $p$ on $T$.

---

References

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