# Topology - Bases of topologies

Prove that each of the following are bases for topologies on the prescribed sets.

This is what I have done thus far on the problem. Each defined topology is listed with my explanation:

In order to do so, we will employ Proposition 4:

Let $X$ be any set. Assume ℘ to be a family of subsets of X such that:

(i) $X$ is the union of members of ℘;

(ii) The intersection of any two members of ℘ is the union of members of ℘.

The topology is defined to be the union of members of ℘. ℘ is a basis for the topology.

a) The set of intervals of the form $[a, b)$ in the set of real numbers, which we define to be $B_i$.

Proof: Here, our set $X$ is defined to be the set of real numbers, and we must show that the set of intervals of the form $[a, b)$ form a basis ℘ for the topology on $X$.

Let us first show that the real numbers can be represented as the union of members of $B_i$, then that the intersection of any two members of $B_i$ can be represented as the union of members of $B_i$.

The set of real numbers can be represented as the union of all intervals of the form $[a_i, b_j)$ such that $a_i$ decreases to negative infinity and $b_j$ increases to positive infinity. If two intervals overlap, their intersection is of the form $[a, b)$ and can, consequently, be constructed by the union of members of $B_i$. If two intervals do not overlap, then this yields the empty set, which can be obtained by taking the union of the empty set with itself. Thus, we have shown that the set Bi is a basis for the topology on the set of real numbers.

b) $X := \{ f | f$ is a function from $[0, 1]$ into $[0, 1]\}$ and the collection of subsets of $X$ of the form

$B_s$ = $\{ f$ is an element of $X | f(x)=0$ for any $x$ that is an element of $S \}$

Proof: Here, our set $X$ is defined as above, and we must show that $B_s$ forms a basis ℘ for the topology on $X$. We will prove this similarly as we did in (4a).

The set $X$ is the union of members of $B_s$ because we can define $S$ as needed so that every point in the image is “hit” by the function, e.g. if $S=0$ or $S > 0$.

Intuitively, with this definition, we can construct the topology with $B_s$ and, furthermore, the topology is a basis of itself. Thus, we have shown that the set $B_s$ is a basis for the topology on the set X.

c) $X := \{ p | p$ is a polynomial with real coefficients$\}$ and the collection of subsets of $X$ of the form

$B_n$ = $\{ p$ is an element of $X |$ degree of $p = n \}$, where $n$ is a nonnegative integer

Similar to (4a), $X$ (the set of all polynomials with real coefficients) can be represented by the union of all polynomials represented by $B_n$.

In this case, $B_n$ is disjoint and our condition (ii) is automatically satisfied since the intersection of any two polynomials represented by $B_n$ is the empty set. The union between the empty set and any other polynomial remains in the form defined by $B_n$. Thus, we have shown that the set $B_n$ is a basis for the topology on the set $X$.

Am I going in the right direction? Please, let me know if there is anything further that I should do to any of the above.

Your proof for a) is essentially correct: it's pretty clear that e.g. $\mathbb{R} = \cup_{n \in \mathbb{Z}} [n, n+1)$, so we can write the whole space as a union of elements from the family $\mathcal{B}$ (the supposed base), and every intersection of two such elements is either of the same form (and so trivially a union (of one) from this family) or empty (and thus equal to the empty union, i.e. the union with no elements) (or note that we could consider $\emptyset = [0, -1)$ to be part of the collection from the start...).

For b) you start to become vague. Be exact! You claim that $X$ equals the union of all sets $B_S$, where $S \subset [0,1]$. So let $f \in X$. Define $S(f) = \{x \in [0,1]: f(x) = 0\}$, then by definition (!) $f \in B_{S(f)}$, and you are done with the first part of the criterion. Alternatively, note that $B_{\emptyset} = \{f: \forall x \in \emptyset: f(x) = 0 \} = X$, because a universal statement over the empty set is always true, and we wrote $X$ as a one set union, a member of the family.

For the second, note (easy proof, do it!) that $B_S \cap B_T = B_{S \cup T}$, for all $S,T \subset [0,1]$, so like for a), the collection is closed under intersections of two elements, which is even more than required.

For c): every polynomial $p$ has a finite degree $n$ which makes it a member of $B_n$. And any two different $B_n$ have empty intersection (the $B_n$ form a partition of the space), because a polynomial has a unique degree. So their intersection is an empty union again! And this is really all for that part, as you already did. No need to talk about unions in the second part.

So all in all, you did two of them right, and were vague on the details of b).

We can rewrite condition 2 by

$\forall x\in U\cap V\exists W(x\in W\wedge W\subseteq U\cap V)$, where $U,V,W$ are basis sets.

a) We just need to prove there is always a interval sitting in the intersection of two intervals.

b) Suppose $f\in B_{S_1}\cap B_{S_2}$, just take your $W$ to be $B_{S_1}\cap B_{S_2}$ because it's already a basis set, since $B_{S_1}\cap B_{S_2}=B_{S_1\cap S_2}$.

c) All $B_n$ are mutually disjoint.

• Could you explain (b) a bit further please? – kathystehl Feb 4 '14 at 19:56
• @SKA For two it suffices to show the intersection of two basis sets is a basis sets. Let $B_U,B_V$ be two basis sets, the intersection of them consists of function that is $0$ on both $U$ and $V$, i.e., $B_U\cap B_V=B_{U\cap V}$. – Kaa1el Feb 5 '14 at 3:04
• @SKA Actually for a) you can do the same. These are all trivial examples of topological basis. – Kaa1el Feb 5 '14 at 3:07

For the B case I have the following solution. am I right?

Given the unit interval $I=[0,1]\subset \mathbb{R}$, let $$X= \text{MOR}(I,I)=\{f|f:I\to I\}$$ for each $S\subseteq I$ we define $$B_S=\{f\in X:f(x)=0\ \ \ \text{for each}\ \ \ x\in S\}$$

Let see that $\beta=\{B_S\}$ is a base.

• Since $S\subseteq I$ them $S\in \wp (I)$ we have that $$\bigcup_{S\in\wp (I)}B_S=X=\text{MOR}(I,I)$$
• Let $R\in B_S\cap B_T$ them $f \in B_S$ y $f \in B_T$ them we have $f(x)=0$ for each $x\in S$ and $x\in T$. If $R=S\cap T$ them $f(x)=0$ for all $x\in R$, them $$B_R \subset B_S\cap B_T$$ and $$f\in B_R \subset B_S\cap B_T$$

Them we have that $\beta$ is base for a topology