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Here is my attempt at showing this collection forms a basis. Have I done all the necessary cases? Is this correct?

Problem:We construct a topological space. Let $\mathbb{R}_{+00}$ be the set consisting of $\mathbb{R}_+$ together with two points called $0'$ and $0''$. Put a topology on it generated by a basis consisting of all intervals in $\mathbb{R}_+$ of the form $(a,b)$ or else of the form $(0,b) \cup \{0'\}$ or $(0,b) \cup \{0''\}$, $a,b \in \mathbb{R}_+$. Show this collection of sets forms a basis for a topology.

Attempt: Each point in $\mathbb{R}_{+00}$ is in at least one basis element. If $x \in \mathbb{R_+}$,the basis element $(0,x+1) \cup \{0'\}$ certainly contains $x$. $0' \in (0,b) \cup \{0'\}$ for any $b \in \mathbb{R_+}$ and $0'' \in (0,b) \cup \{0''\}$ for any $b \in \mathbb{R_+}$. The intersection of any two open intervals $(a,b),(c,d),a,b,c,d \in \mathbb{R_+},a<b,c<d$ is either $\varnothing$, or a basis element. If $a<b$ and $x \in ((0,a) \cup \{0'\}) \cap ((0,b) \cup \{0''\})$, choose $\epsilon$ such that $0<\epsilon<x$. Then $x\in (x-\epsilon,a) \subset ((0,a) \cup \{0'\}) \cap ((0,b) \cup \{0''\})$. The case where $b>a$ is similar. $(c,d) \cap ((0,a) \cup \{0'\})$ is either $\varnothing$ or an open interval of the form $(e,f)$, a basis element. Similarly for $(c,d) \cap ((0,a) \cup \{0''\}).$

Scott Frazier
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  • Looks nice for me. –  Jan 30 '21 at 04:06
  • Are we supposed to assume $0',0''$ are distinct negative points? – Shubham Johri Jan 30 '21 at 04:15
  • @ShubhamJohri I am not sure, it just says two made up points. – Scott Frazier Jan 30 '21 at 04:18
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    @ShubhamJohri: No, they are simply two new points not in $\Bbb R_+$. The topology generated by this base makes each of the subsets $\Bbb R_+\cup{0'}$ and $\Bbb R_+\cup{0''}$ homeomorphic to $\Bbb R_{\ge 0}$ with the usual topology; the space as a whole is the line with two origins minus the negative reals. – Brian M. Scott Jan 30 '21 at 04:37
  • @ScottFrazier If you don't know that $0',0''\le0$, how does your second last line make sense? – Shubham Johri Jan 30 '21 at 04:45
  • If $B$ is a family of subsets of a set $X$ then $B$ is a base (basis) for a topology on $X$ iff $(i) ,\cup B=X,$(i.e. every member pf $X$ belongs to at least one $b\in B$), and $(ii)$ whenever $b,b'\in B$ and $x\in b\cap b',$ there exists $b''\in B$ with $x\in b''\subseteq b\cap b'.$ Of course if $b\cap b'\in B$ whenever $b,b'\in B$ then $(ii)$ is satisfied by $b''=b\cap b'$. – DanielWainfleet Jan 30 '21 at 05:46
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    @DanielWainfleet I do not mean to be rude, but I thought those conditions are what I have shown.Where did I go wrong? At least that is what I have tried to show. – Scott Frazier Jan 30 '21 at 05:49
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    You aren't wrong. I just wanted to state the general case. Some other students get the mistaken notion that a base must be closed under finite intersections – DanielWainfleet Jan 30 '21 at 06:13

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Without knowing the sign of $0',0''$, your assumption that $(c,d)\cap[(0,a)\cup\{0'\}]$ being either empty or an open interval is incorrect. Note that it is equal to$$A=[(c,d)\cap(0,a)]\color{pink}\cup[(c,d)\cap\{0'\}]=(\max\{0,c\},\min\{a,d\})\color{pink}\cup[(c,d)\cap\{0'\}]$$where the former is an open interval (including the empty set) and the latter may give you an isolated point $0'$ in case $0'$ is positive. In this case there is no basis element containing $0'$ which is a subset of $A$ unless $c=0$. For example, $(c,d)=(5,15),(0,a)\cup\{0'\}=(0,6)\cup\{12\}$.

So you must remark that the given set forms a basis iff $0',0''\le0$.

For $a<b$,$$[(0,a) \cup \{0'\}] \cap[(0,b) \cup \{0''\}]=(0,a)\color{pink}\cup[\{0'\}\cap\{0''\}]$$There is a possibility that this intersection contains the point $0'$ iff $0'=0''$, in which case $(0'-\epsilon,a)$ is not be a suitable choice. For $0'$, select the basis element $(0,a)\cup\{0'\}$ and you are done.

Shubham Johri
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