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# 数学代写|泛函分析代写Functional Analysis代考|Topological Properties of Metric Spaces

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## 数学代写|泛函分析代写FUNCTIONAL ANALYSIS代考|Topological Properties of Metric Spaces

Let $X=(X, d)$ be a metric space. Defining, for every $x \in X$, the family $\mathcal{B}_x$ of neighborhoods of $x$ as the family of open balls centered at $x$
$$\mathcal{B}_x={B(x, \varepsilon), \varepsilon>0}$$
we introduce in $X$ a topology induced by the metric $d$. Thus every metric space is a topological space with the topology induced by the metric. Two immediate corollaries follow:
(i) Bases $\mathcal{B}_x$ are of countable type.
(ii) The metric topology is Hausdorff.
The first observation follows from the fact that $\mathcal{B}_x$ is equivalent to its subbase of the form
$$\left{B\left(x, \frac{1}{k}\right), \quad k=1,2, \ldots\right}$$
To prove the second assertion consider two distinct points $x \neq y$. We claim that balls $B(x, \varepsilon)$ and $B(y, \varepsilon)$, where $\varepsilon=d(x, y) / 2$, are disjoint. Indeed, if $z$ were a point belonging to the balls simultaneously, then
$$d(x, y) \leq d(x, z)+d(z, y)<\varepsilon+\varepsilon=d(x, y)$$
Thus all the results we have derived in the first five sections of this chapter for Hausdorff first countable topological spaces hold also for metric spaces. Let us briefly review some of them.

## 数学代写|泛函分析代写FUNCTIONAL ANALYSIS代考|Open and Closed Sets in Metric Spaces

Open and Closed Sets in Metric Spaces. A set $G \subset X$ is open if and only if, for every point $x$ of $G$, there exists a ball $B(x, \varepsilon)$, centered at $x$, that is contained in $G$. A point $x$ is an accumulation point of a set $F$ if every ball centered at $x$ contains points from $F$ which are different from $x$, or, equivalently, there exists a sequence $x_n$ points of $F$ converging to $x$.
Note that a sequence $x_n$ converges to $x$ if and only if
$$\forall \varepsilon>0 \exists N=N(\varepsilon): d\left(x_n, x\right)<\varepsilon \quad \forall n \geq N$$
Finally, a set is closed if it contains all its accumulation points.
Continuity in Metric Spaces. Let $(X, d)$ and $(Y, \rho)$ be two metric spaces. Recall that a function $f: X \rightarrow$ $Y$ is continuous at $x_0$ if
$$f\left(\mathcal{B}{x_0}\right) \succ \mathcal{B}{f\left(x_0\right)}$$
or, equivalently,
$$\forall \varepsilon>0 \quad \exists \delta>0: f\left(B\left(x_0, \delta\right)\right) \subset B\left(f\left(x_0\right), \varepsilon\right)$$
The last condition can be put into a more familiar form of the definition of continuity for metric spaces $(\varepsilon-\delta$ continuity):
Function $f: X \rightarrow Y$ is continuous at $x_0$ if and only if for every $\varepsilon>0$ there is a $\delta=\delta\left(\varepsilon, x_0\right)$ such that
$$\rho\left(f(x), f\left(x_0\right)\right)<\varepsilon \quad \text { whenever } \quad d\left(x, x_0\right)<\delta$$ Note that number $\delta$ generally depends not only on $\varepsilon$, but also upon the choice of point $x_0$. If $\delta$ happens to be independent of $x_0$ for all $x_0$ from a set $E$, then $f$ is said to be uniformly continuous on $E$. Let us recall also that, since bases of neighborhoods are of countable type, i.e., metric spaces are first countable topological spaces, continuity in metric spaces is equivalent to sequential continuity: a function $f: X \rightarrow Y$ is continuous at $x_0$ if and only if $$f\left(x_n\right) \rightarrow f\left(x_0\right) \quad \text { whenever } \quad x_n \rightarrow x_0$$ Suppose now that there exists a constant $C>0$, such that
$$\rho(f(x), f(y)) \leq C d(x, y) \quad \text { for every } x, y \in E$$

## 数学代写|泛函分析代写FUNCTIONAL ANALYSIS代考|Topological Properties of Metric Spaces

$$\mathcal{B}_x={B(x, \varepsilon), \varepsilon>0}$$

(i)基数$\mathcal{B}_x$为可数型。
(ii)度量拓扑是Hausdorff。

$$\left{B\left(x, \frac{1}{k}\right), \quad k=1,2, \ldots\right}$$

$$d(x, y) \leq d(x, z)+d(z, y)<\varepsilon+\varepsilon=d(x, y)$$

## 数学代写|泛函分析代写FUNCTIONAL ANALYSIS代考|Open and Closed Sets in Metric Spaces

$$\forall \varepsilon>0 \exists N=N(\varepsilon): d\left(x_n, x\right)<\varepsilon \quad \forall n \geq N$$ 最后，如果一个集合包含了它所有的累加点，那么它就是封闭的。 度量空间的连续性。设$(X, d)$和$(Y, \rho)$是两个度量空间。回想一下，函数$f: X \rightarrow$$Y在x_0 if处是连续的$$ f\left(\mathcal{B}{x_0}\right) \succ \mathcal{B}{f\left(x_0\right)} $$或者，等价地，$$ \forall \varepsilon>0 \quad \exists \delta>0: f\left(B\left(x_0, \delta\right)\right) \subset B\left(f\left(x_0\right), \varepsilon\right) $$最后一个条件可以用更熟悉的度量空间连续性定义形式(\varepsilon-\delta continuity)表示: 函数f: X \rightarrow Y在x_0是连续的当且仅当对于每个\varepsilon>0有一个\delta=\delta\left(\varepsilon, x_0\right)使得$$ \rho\left(f(x), f\left(x_0\right)\right)<\varepsilon \quad \text { whenever } \quad d\left(x, x_0\right)<\delta $$请注意，数字\delta一般不仅取决于\varepsilon，还取决于点x_0的选择。如果对于一个集合E中的所有x_0, \delta恰好独立于x_0，那么f在E上是一致连续的。我们还回顾一下，由于邻域的基是可数型的，即度量空间是第一可数的拓扑空间，因此度量空间中的连续性等价于顺序连续性:一个函数f: X \rightarrow Y在x_0处连续当且仅当$$ f\left(x_n\right) \rightarrow f\left(x_0\right) \quad \text { whenever } \quad x_n \rightarrow x_0 $$，现在假设存在一个常数C>0，使得$$ \rho(f(x), f(y)) \leq C d(x, y) \quad \text { for every } x, y \in E$\$

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