429: 2 Continuous Maps into Hausdorff Topological Space That Disagree at Point Disagree on Neighborhood of Point

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A description/proof of that 2 continuous maps into Hausdorff topological space that disagree at point disagree on neighborhood of point

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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 Hausdorff topological space.
• The reader knows a definition of continuous map.
• The reader knows a definition of neighborhood of point.

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Target Context

• The reader will have a description and a proof of the proposition that any 2 continuous maps from any topological space into any Hausdorff topological space that (the maps) disagree at any point disagree on a neighborhood of the point.

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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 topological space, $T_1$, any Hausdorff topological space, $T_2$, and any continuous maps, $f_1,f_2:T_1\to T_2$, if $f_1(p)\ne f_2(p)$ at any point, $p\in T_1$, there is a neighborhood, $N_p$, of $p$ such that $f_1(p^{\prime })\ne f_2(p^{\prime })$ for each point, $p^{\prime }\in N_p$.

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2: Proof

Let us suppose that $f_1(p)\ne f_2(p)$. There are some disjoint open neighborhoods, $U_{f_1(p)},U_{f_2(p)}\in T_2$, of $f_1(p),f_2(p)$, respectively, such that $U_{f_1(p)}\cap U_{f_2(p)}=\mathrm{\varnothing }$, because $T_2$ is Hausdorff. As $f_i$ is continuous, there is a neighborhood, $N_{p,i}$, of $p$ such that $f_i(N_{p,i})\subseteq U_{f_i(p)}$. $N_p:=N_{p,1}\cap N_{p,2}$ is a neighborhood, and $f_i(N_p)\subseteq U_{f_i(p)}$. $f_1(N_p)\cap f_2(N_p)=\mathrm{\varnothing }$, which means that $f_1(p^{\prime })\ne f_2(p^{\prime })$ for each $p^{\prime }\in N_p$.

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References

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