# Basis of topological space and topological subspace

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 $$\tilde{I}:=\{(a,b)\;|\;-\infty The family $$\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 $$\overline{I}:=\{(a,b)\;|\;-\infty The family $$\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 $$B(x_0,r)=\{x\in\mathbb{R}^n\;|\;d(x_0,x), and $$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!

## 1 Answer

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 $$a. Then letting $$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 $$a. Thus, setting $$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 $$a_{\alpha} for each $$\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 $$a then $$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

$$W\cap\mathbb{R}=(a,\infty)=\bigcup_{n\in\mathbb{N}}(a,a+n)$$

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.

• @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? – Jack J. Nov 4 '18 at 9:21
• I will edit my answer with some help. – Robert Thingum Nov 4 '18 at 20:39
• I appreciate your patience – Jack J. Nov 4 '18 at 20:40
• Let me know if you have questions with my edited answer. – Robert Thingum Nov 4 '18 at 20:57
• That is correct. – Robert Thingum Nov 9 '18 at 15:45