This topics is a continuation of this Basis of topological space. I rewrite the most important theorem:

Theorem. Let $(X, \mathcal{G})$ a topological space.

$(i)$ Let $\mathcal{B}\subseteq\mathcal{G}$ a basis of $(X,\mathcal{G})$. Then,

$(a)$ $\mathcal{B}$ is a coverage of $X$;

$(b)$ for each $B_1,B_2\in\mathcal{B}$, $B_1\cap B_2\ne\emptyset$ and for each $x\in B_1\cap B_2$ exists $B\in\mathcal{B}$ such that $x\in B\subseteq B_1\cap B_2$.

$(ii)$ Let $\mathcal{B}\subseteq \mathcal{P}(X)$ a family of nonempty set for which they are valid $(a)$ and $(b)$, then exists a unique topology $\mathcal{G}$ on $X$ of which $\mathcal{B}$ is basis.


Proposition 1. Let the family set $$ \tilde{I}:=\{(a,b)\;|\;-\infty<a<b<+\infty\}\subseteq\mathcal{P}(\mathbb{R}). $$ The family $\tilde{I}$ is a basis of a single topology on $\mathbb{R}$.

proof. This proof was addressed in the previous topics. Verifying the properties $(a)$ and $(b)$ of the theorem.

The topology on $\mathbb{R}$ of which $\tilde{I}$ is basis we denote it with $\tau(\mathbb{R})$.


Proposition 2. Let the family set $$\overline{I}:=\{(a,b)\;|\;-\infty<a<b<+\infty\}\cup\{[-\infty,b)\;|\;-\infty<b<+\infty\}\cup\{(a,+\infty]\;|\;-\infty<a<+\infty\}\subseteq\mathcal{P}(\mathbb{\overline{R}}).$$ The family $\overline{I}$ is a basis of a single topology on $\mathbb{\overline{R}=\mathbb{R}\cup{\{\pm\infty\}}}$.

proof. If $x\ne\pm\infty$, we proceed as in the Proposition 1. Otherwise if $x=+\infty$, we consider $(x-\epsilon,+\infty]$, where $\epsilon>0$. In the same way if $x=-\infty$. Therefore the property $(a)$ of the theorem is verified. How can I proceed to check point $(b)$?

The topology on $\mathbb{\overline{R}}$ of which $\overline{I}$ is basis we denote it with $\tau(\mathbb{\overline{R}})$.


Proposition 3. Let the family $$V=\{B(x_0,r)\;|\;x_0\in X,r>0\}\subseteq\mathcal{P}(\mathbb{R}^n),$$ where $B(x_0,r)=\{x\in\mathbb{R}^n\;|\;d(x_0,x)<r\}$, and $d$ is the Euclidean distance.

proof If $x\in\mathbb{R}^n$, we consider $B(x,r)$. Then $x\in B(x,r)$, therefore the property $(a)$ is verified. How can I proceed to check point $(b)$?

Another question: why $\tau(\mathbb{\overline{R}})\cap \mathbb{R}=\tau(\mathbb{R})$? From here how can I conclude that the topological space $(\mathbb{R},\tau(\mathbb{R}))$ is a topological subspace of $(\overline{\mathbb{R}},\tau(\mathbb{\overline{R}})$?

Thanks in advance!


For Proposition two you will have to break the proof into cases. Given the candidate basis $\bar{I}$ you have described you need to consider $U,V\in\bar{I}$ and $x\in U\cap V$. You need to show that there is some $W\in\bar{I}$ such that $x\in W\subseteq U\cap V$. However, you have several different possibilities for the nature of $U$ and $V$.

Case 1: $U,V\in\tilde{I}$. In this case $U$ and $V$ are both bounded open intervals and you say that you've shown that there if $x\in U\cap V$ then there is another open interval $W\in\tilde{I}$ such that $x\in W\subseteq U\cap V$. Because $\tilde{I}\subseteq\bar{I}$ there is then nothing to be done.

Case 2: $U\in\tilde{I}$ and $V\in\bar{I}\setminus\tilde{I}$. Say without loss of generality that $U=(a,b)$ and $V=(c,\infty]$ (the case where $V=[-\infty,c)$ is similar and we won't consider it here). If $x\in U\cap V$ then $a<x<b<\infty$. Then letting $d=\max\{a,c\}$ we have that $x\in(d,b)$. Setting $W=(d,b)$ we see that $x\in W\subseteq U\cap V$ and $W\in\bar{I}$.

Case 3: $U,V\in\bar{I}\setminus\tilde{I}$. Technically this case splits into two further cases but we will only consider the case where $U=[-\infty,b)$ and $V=(a,\infty]$ for the simple reason that I don't want to write out the other case. Now if $x\in U\cap V$ then $a<x<b$. Thus, setting $W=(a,b)$ we have that $W\in\bar{I}$ and $x\in W\subseteq U\cap V$.

I'm assuming that you can fill in the cases that I ignored.

Regarding Prop 3 you will have to wield the triangle inequality.

Let $y,z\in\mathbb{R}^{n}$ and $\epsilon,\delta>0$. Define $U=B(y,\epsilon)$ and $V=B(z,\delta)$. Assume that $x\in U\cap V$. Then, by definition $d(x,y)<\epsilon$ and $d(x,z)<\delta$. Say $\eta_{1}=\epsilon-d(x,y)$ and $\eta_{2}=\delta-d(x,z)$. Define $\xi=\frac{1}{2}\min\{\eta_{1},\eta_{2}\}$. We claim that $W:=B(x,\xi)$ is such that $x\in W\subseteq U\cap V$. It is clear that $x\in W$. Assume that $x^{\prime}\in W$. Then $d(x,x^{\prime})<\xi$. By the triangle inequality we then have that

$$d(x^{\prime},y)\leq\eta_{1}+\xi<\epsilon$$ $$d(x^{\prime},z)\leq\eta_{2}+\xi<\delta$$

Thus $x^{\prime}\in U\cap V$.

I'm not going to expound on your last question. I will simply give you the hint that you know that $\mathcal{T}(\mathbb{R})\subseteq\{U\cap\mathbb{R}\mid U\in\mathcal{T}(\bar{\mathbb{R}})\}$. You then need only show that if $U\in\mathcal{T}(\bar{\mathbb{R}})$ then you need to show that you can write $U\cap\mathbb{R}$ as a union of bounded open intervals.

Let's now show that $\mathcal{T}(\mathbb{R})=\mathcal{T}(\bar{\mathbb{R}})\cap\mathbb{R}$. We will do this directly.

Let $U\in\mathcal{T}(\mathbb{R})$. Then $U=\bigcup_{\alpha\in A}(a_{\alpha},b_{\alpha})$ where $a_{\alpha}<b_{\alpha}$ for each $\alpha\in A$. That is $U$ is a union of elements of $\tilde{I}$. Because $\tilde{I}\subseteq\bar{I}\subseteq\mathcal{T}(\bar{\mathbb{R}})$ we have that $U\in\mathcal{T}(\bar{\mathbb{R}})$. Now we can determine:

$$U\cap\mathbb{R}=\left(\bigcup_{\alpha\in A}(a_{\alpha},b_{\alpha})\right)\cap\mathbb{R}=\bigcup_{\alpha\in A}[(a_{\alpha},b_{\alpha})\cap\mathbb{R}]=\bigcup_{\alpha\in A}(a_{\alpha},b_{\alpha})=U$$

Therefore $U\in\mathcal{T}(\bar{\mathbb{R}})\cap\mathbb{R}$ and hence $\mathcal{T}(\mathbb{R})\subseteq\mathcal{T}(\bar{\mathbb{R}})\cap\mathbb{R}$.

Now we will show that $\mathcal{T}(\bar{\mathbb{R}})\cap\mathbb{R}\subseteq\mathcal{T}(\mathbb{R})$. To do this we, as you did in the comments, take $V\in\mathcal{T}(\bar{\mathbb{R}})$. Noting that $V=\bigcup_{\beta\in B}W_{\beta}$ where $W_{\beta}\in\bar{I}$ for each $\beta$. Now, note that we have the following:

$$V\cap\mathbb{R}=\left(\bigcup_{\beta\in B}W_{\beta}\right)\cap\mathbb{R}=\bigcup_{\beta\in B}(W_{\beta}\cap\mathbb{R})$$

Because $\mathcal{T}(\mathbb{R})$ is closed under arbitrary unions it will then suffice to show that $W\cap\mathbb{R}\in\mathcal{T}(\mathbb{R})$ for each $W\in\bar{I}$ (Exercise: Why is that?). We then consider our various cases. If $W=(a,b)$ where $a,b\in\mathbb{R}$ and $a<b$ then $W\cap\mathbb{R}=W$ which is an element of $\tilde{I}$ which is itself a subset of $\mathcal{T}(\mathbb{R})$ yielding $W\in\mathcal{T}(\mathbb{R})$. We then consider the case where $W=(a,\infty]$ for some $a\in\mathbb{R}$. In this case we have


Each open interval $(a,a+n)$ is in $\tilde{I}\subseteq\mathcal{T}(\mathbb{R})$, so $W\cap\mathbb{R}=\bigcup_{n\in\mathbb{N}}(a,a+n)$ is an element of $\mathcal{T}(\mathbb{R})$. The final case is where $W=[-\infty,a)$ where $a\in\mathbb{R}$, however this case is very similar to the previous one.

We then have that $W\cap\mathbb{R}\in\mathcal{T}(\mathbb{R})$ for each $W\in\bar{I}$. Therefore if $V=\bigcup_{\beta\in B}W_{\beta}$ is an element of $\mathcal{T}(\bar{\mathbb{R}})$ where $W_{\beta}\in\bar{I}$ for each $\beta$ then as shown above $V\cap\mathbb{R}=\bigcup_{\beta\in B}(W_{\beta}\cap\mathbb{R})$. We just showed that each $W_{\beta}\cap\mathbb{R}$ is an element of $\mathcal{T}(\mathbb{R})$ and $\mathcal{T}(\mathbb{R})$ is closed under arbitrary unions so $V\cap\mathbb{R}$ is an element of $\mathcal{T}(\mathbb{R})$. Therefore, $\mathcal{T}(\bar{\mathbb{R}})\cap\mathbb{R}\subseteq\mathcal{T}(\mathbb{R})$, establishing equality.

Apologies for being verbose, but I wanted to be thorough.

  • $\begingroup$ @RobertThingumThanks for your answer. If $U\in\tau({\overline{\mathbb{R}}})$, then $U=\cup_{i\in I}B_i$, where $\{B_i\}\subseteq\overline{I}$. Therefore $U\cap\mathbb{R}=\cup_{i\in I}(B_i\cap\mathbb{R})$. Then $\tau(\mathbb{R})\supseteq\{U\cap\mathbb{R}\;|\;U\in\tau(\overline{\mathbb{R}})\}$. Therefore $\tau(\mathbb{R})=\{U\cap\mathbb{R}\;|\;U\in\tau(\overline{\mathbb{R}})\}:=\tau(\overline{\mathbb{R}})\cap\mathbb{R}$. Correct? $\endgroup$ – Jack J. Nov 4 '18 at 9:21
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    $\begingroup$ I will edit my answer with some help. $\endgroup$ – Robert Thingum Nov 4 '18 at 20:39
  • $\begingroup$ I appreciate your patience $\endgroup$ – Jack J. Nov 4 '18 at 20:40
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    $\begingroup$ Let me know if you have questions with my edited answer. $\endgroup$ – Robert Thingum Nov 4 '18 at 20:57
  • 1
    $\begingroup$ That is correct. $\endgroup$ – Robert Thingum Nov 9 '18 at 15:45

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