248: For Map Between Real Closed Intervals and Graph of Map as Topological Subspace, Subset Such That Value Is Larger or Smaller Than Independent Variable Is Open

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A description/proof of that for map between real closed intervals and graph of map as topological subspace, subset such that value is larger or smaller than independent variable is open

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Topics

About: topological space

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The table of contents of this article

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

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Starting Context

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

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Target Context

• The reader will have a description and a proof of the proposition that for any map between any real closed intervals and the graph of the map as topological subspace, the subset such that value is larger than independent variable and the subset such that value is smaller than independent variable are open.

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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.

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Main Body

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1: Description

For any real closed intervals, $[r_1,r_2]$ and $[r_3,r_4]$, any map, $f:[r_1,r_2]\to [r_3,r_4]$, and the graph of the map, $T=\{(r,f(r))\in [r_1,r_2]\times [r_3,r_4]\}$ with the subspace topology of $[r_1,r_2]\times [r_3,r_4]$ where $[r_1,r_2]\times [r_3,r_4]$ is with the subspace topology of $\mathbb{R}\times \mathbb{R}$, $S_1=\{(r,f(r))\in T|r<f(r)\}$ and $S_2=\{(r,f(r))\in T|f(r)<r\}$ are open on $T$.

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2: Proof

Let us think of $S_1$. Its openness is about the existence of a $T$ open set, $U_p\subset T$, around any point, $p=(r,f(r))\in S_1$, such that $U_p\subseteq S_1$. It suffices to show that there is an open ball, $B_{p-ϵ}\subseteq \mathbb{R}\times \mathbb{R}$ such that $B_{p-ϵ}\cap T\subseteq S_1$, because $B_{p-ϵ}\cap ([r_1,r_2]\times [r_3,r_4])$ is open on $[r_1,r_2]\times [r_3,r_4]$ by the definition of subspace topology and $B_{p-ϵ}\cap T=B_{p-ϵ}\cap ([r_1,r_2]\times [r_3,r_4])\cap T$ is open on $T$ by the definition of subspace topology. It is suffice to show that there is a $B_{p-ϵ}$ that is in the internal area over the line that goes through $(r_1,r_3)$ and $(r_1+1,r_3+1)$ ("internal" means that the line is not included), because on the area, $r<f(r)$ holds, so it holds on $B_{p-ϵ}$, so, for any point, $p^{\prime }\in B_{p-ϵ}\cap T$, $p^{\prime }\in S_1$. Such a $B_{p-ϵ}$ indeed exists because $p$ is not on the line, so, the distance from $p$ to the line is a positive, $d$, and any $ϵ<d$ suffices.

$S_2$ is likewise.

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References

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