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# 数学代写|拓扑学代写TOPOLOGY代考|Sub-bases and Alexander’s Theorem

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## 数学代写|拓扑学代写TOPOLOGY代考|Sub-bases and Alexander’s Theorem

Definition 7.1 A sub-basis of a topological space is a family $\mathcal{P}$ of open sets such that finite intersections in $\mathcal{P}$ form a basis of the topology.
Every basis is also a sub-basis.
Example 7.2 A sub-basis for the Euclidean topology on $\mathbb{R}$ is given by the open sets ]$-\infty, a[] b,,+\infty[$ as $a, b \in \mathbb{R}$ vary, since open intervals $] a, b[$ form a basis, and we can write $] a, b[=]-\infty, b[\cap] a,+\infty[$.

Lemma 7.3 Let $X, Y$ be topological spaces and $\mathcal{P}$ a sub-basis of $Y$. A map $f: X \rightarrow$ $Y$ is continuous if and only if $f^{-1}(U)$ is open for every $U \in \mathcal{P}$.
Proof Just observe that $f^{-1}$ commutes with union and intersection.
Let $\mathcal{P}$ be a cover of a set $X$, and consider the family $\mathcal{B}$ of finite intersections in $\mathcal{P}$. If $A, B \in \mathcal{B}$ then $A \cap B \in \mathcal{B}$, and $\mathcal{B}$ covers $X$. By Theorem $3.7 \mathcal{B}$ is a basis of a topology that has $\mathcal{P}$ as sub-basis. It’s easy to see that this topology is the coarsest one having $\mathcal{P}$ as open sets.

Example 7.4 Take a set $S$ and a topological space $X$; in the set $X^S$ of maps $f: S \rightarrow X$ consider the family $\mathcal{P}$ of subsets
$$P(s, U)={f: S \rightarrow X \mid f(s) \in U}$$
for any $s \in S$ and any open set $U$. The coarsest topology on $X^S$ containing $\mathcal{P}$ is called pointwise-convergence topology, and has $\mathcal{P}$ as sub-basis.

## 数学代写|拓扑学代写TOPOLOGY代考|Infinite Products

Given an arbitrary family $\left{X_i \mid i \in I\right}$ of sets one defines the Cartesian product $X=\prod_{i \in I} X_i$ as the set of all maps $x: I \rightarrow \cup_i X_i$ such that $x_i \in X_i$ for every index $i \in I$. That’s to say that any element of the product is a collection $\left{x_i\right}_{i \in I}$ indexed by $I$ such that $x_i \in X_i$ for every $i$. The axiom of choice ensures that $X$ isn’t empty provided each $X_i$ is non-empty. The projections $p_i: X \rightarrow X_i$ are defined as $p_i(x)=x_i$

When the $X_i$ are all equal to some set $X_0$, the product $\prod_{i \in I} X_i=\prod_{i \in I} X_0$ coincides with $X_0^I$, the set of maps $I \rightarrow X_0$. For any set $Y$ and any $f: Y \rightarrow$ $\prod_{i \in I} X_i$ we write $f_i$ to mean $f$ followed by the projection $p_i$. Note that $f$ is uniquely determined by the family $\left{f_i: Y \rightarrow X_i \mid i \in I\right}$.

If all $X_i$ are topological spaces, we define product topology on $X$ the coarsest one for which the projections are continuous. This amounts to say that $p_i^{-1}(U)$, for every $i \in I$ and $U$ open in $X_i$, constitute a sub-basis that we shall call canonical sub-basis. A canonical basis is given by open sets that are finite intersections of the canonical sub-basis.

When the $X_i$ are all equal the product topology coincides with the pointwiseconvergence topology (Example 7.4), because the two have the same sub-basis.
Lemma 7.6 In the previous notations, a map $f: Y \rightarrow \prod_{i \in I} X_i$ is continuous if and only if all its components $f_i: Y \rightarrow X_i$ are continuous.

Proof The projections $p_i$ are continuous, so $f$ continuous implies that the components $f_i=p_i f$ are.

Conversely, suppose each component $f_i$ is continuous. For any open set $p_i^{-1}(U)$ in the canonical sub-basis we have $f^{-1}\left(p_i^{-1}(U)\right)=f_i^{-1}(U)$ is open, implying $f$ continuous.

## 数学代写|拓扑学代写TOPOLOGY代考|Sub-bases and Alexander’s Theorem

$$P(s, U)={f: S \rightarrow X \mid f(s) \in U}$$

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