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# 澳洲代考|拓扑学代考Topology代考|MAST90023

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## 澳洲代考|拓扑学代考Topology代考|CONVERGENCE, COMPLETENESS, AND BAIRE’S THEOREM

As we emphasized in the introduction to this chapter, one of our main aims in considering metric spaces is to study convergent sequences in a context more general than that of classical analysis. The fruits of this study are many, and among them is the added insight gained into ordinary convergence as it is used in analysis.
Let $X$ be a metric space with metric $d$, and let
$$\left{x_{n}\right}=\left{x_{1}, x_{2}, \ldots, x_{n}, \ldots .\right}$$
be a sequence of points in $X$. We say that $\left{x_{n}\right}$ is convergent if there exists a point $x$ in $X$ such that either
(1) for each $\epsilon>0$, there exists a positive integer $n_{0}$ such that $n \geq n_{0} \Rightarrow d\left(x_{n}, x\right)<\epsilon$ or equivalently,
(2) for each open sphere $S_{\mathrm{e}}(x)$ centered on $x$, there exists a positive integer $n_{0}$ such that $x_{n}$ is in $S_{\epsilon}(x)$ for all $n \geq n_{0}$.

## 澳洲代考|拓扑学代考Topology代考|CONTINUOUS MAPPINGS

There is some rather undescriptive terminology which is of ten used in connection with Baire’s theorem. We shall not make use of it ourselves, but the reader ought to be acquainted with it. A subset of a metric space is called a set of the first category if it can be represented as the union of a sequence of nowhere dense sets, and a set of the second category if it is not a set of the first category. Baire’s theorem-sometimes called the Baire category theorem-can now be expressed as follows: any complete metric space (considered as a subset of itself) is a set of the second category.
of the following equivalent conditions is satisfied:
(1) for each $\epsilon>0$ there exists $\delta>0$ such that $d_{1}\left(x, x_{0}\right)<\delta \Rightarrow$ $d_{2}\left(f(x), f\left(x_{0}\right)\right)<\epsilon$
(2) for each open sphere $S_{\epsilon}\left(f\left(x_{0}\right)\right)$ centered on $f\left(x_{0}\right)$ there exists an open sphere $S_{\delta}\left(x_{0}\right)$ centered on $x_{0}$ such that $f\left(S_{\delta}\left(x_{0}\right)\right) \subseteq S_{\epsilon}\left(f\left(x_{0}\right)\right)$.
The reader will notice that the first condition generalizes the elementary definition given in the introduction to this chapter, and that the second translates the first into the language of open spheres.

Our first theorem expresses continuity at a point in terms of sequences which converge to the point.

## 澳洲代考|拓扑学代考TOPOLOGY代考|SPACES OF CONTINUOUS FUNCTIONS

Let $L$ be a non-empty set, and assume that each pair of elements $x$ and $y$ in $L$ can be combined by a process called addition to yield an element $z$ in $L$ denoted by $z=x+y$. Assume also that this operation of addition satisfies the following conditions:
(1) $x+y=y+x$;
(2) $x+(y+z)=(x+y)+z$
(3) there exists in $L$ a unique element, denoted by 0 and called the zero element, or the origin, such that $x+0=x$ for every $x$;
(4) to each element $x$ in $L$ there corresponds a unique element in $L$, denoted by $-x$ and called the negative of $x$, such that $x+(-x)=0$.

## 澳洲代考|拓扑学代考TOPOLOGY代考|CONVERGENCE, COMPLETENESS, AND BAIRE’S THEOREM

\left{x_{n}\right}=\left{x_{1}, x_{2}, \ldots, x_{n}, \ldots .\right}\left{x_{n}\right}=\left{x_{1}, x_{2}, \ldots, x_{n}, \ldots .\right}

1对于每个ε>0, 存在一个正整数n0这样n≥n0⇒d(Xn,X)<ε或等效地，
2对于每个开放球体小号和(X)以X, 存在一个正整数n0这样Xn在小号ε(X)对所有人n≥n0.

## 澳洲代考|拓扑学代考TOPOLOGY代考|CONTINUOUS MAPPINGS

1对于每个ε>0那里存在d>0这样d1(X,X0)<d⇒ d2(F(X),F(X0))<ε
2对于每个开放球体小号ε(F(X0))以F(X0)存在一个开放的球体小号d(X0)以X0这样F(小号d(X0))⊆小号ε(F(X0)).

## 澳洲代考|拓扑学代考TOPOLOGY代考|SPACES OF CONTINUOUS FUNCTIONS

1 X+是=是+X;
2 X+(是+和)=(X+是)+和
3存在于大号一个唯一元素，用 0 表示，称为零元素或原点，这样X+0=X对于每个X;
4对每个元素X在大号对应于一个独特的元素大号，表示为−X并称为负X, 这样X+(−X)=0.

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