数学代写|拓扑学代写Topology代考|THE COMPONENTS OF A SPACE

数学代写|拓扑学代写Topology代考|THE COMPONENTS OF A SPACE

If a space is not itself connected, then the next best thing is to be able to decompose it into a disjoint class of maximal connected subspaces. Our present objective is to show that this can always be done.

A maximal connected subspace of a topological space, that is, a connected subspace which is not properly contained in any larger connected subspace, is called a component of the space. A connected space clearly has only one component, namely, the space itself. In a discrete space, it is easy to see that each point is a component.

The following two theorems will be useful in obtaining the desired decomposition for a general space.

Theorem A. Let $X$ be a topological space. If $\left{A_i\right}$ is a non-empty class of connected subspaces of $X$ such that $\cap_i A_i$ is non-empty, then $A=\cup_1 A_i$ is also a connected subspace of $X$.
proof. Assume that $A$ is disconnected. This means that there exist two open subsets $G$ and $H$ of $X$ whose union contains $A$ and whose intersections with $A$ are disjoint and non-empty. All the $A_i$ ‘s are connected, and each lies in $G \cup H$, so each $A_i$ lies entirely in $G$ or entirely in $H$ and is disjoint from the other. Since $\bigcap_i A_i$ is non-empty, either all the $A_i$ ‘s lie in $G$ and are disjoint from $H$, or all lie in $H$ and are disjoint from $G$. We see by this that $A$ itself is disjoint from either $G$ or $H$, and this contradiction shows that our assumption that $A$ is disconnected is untenable.
Theorem B. Let $X$ be a topological space and A a connected subspace of $X$. If $B$ is a subspace of $X$ such that $A \subseteq B \subseteq \bar{A}$, then $B$ is connected; in particular, $\bar{A}$ is connected.
proof. Assume that $B$ is disconnected, that is, that there exist two open subsets $G$ and $H$ of $X$ whose union contains $B$ and whose intersections with $B$ are disjoint and non-empty. Since $A$ is connected and contained in $G \cup H, A$ is contained in either $G$ or $H$ and is disjoint from the other. Let us say, just to be specific, that $A$ is disjoint from $H$. This implies that $\bar{A}$ is also disjoint from $H$, and since $B \subseteq \bar{A}, B$ is disjoint from $H$. This contradiction shows that $B$ cannot be disconnected, and proves our theorem.

数学代写|拓扑学代写Topology代考|LOCALLY CONNECTED SPACES

In Sec. 23 we encountered the concept of a locally compact space, that is, of a space which is compact around each point but need not be compact as a whole. We now study another “local” property which a topological space may have, that of being connected in the vicinity of each of its points.

A locally connected space is a topological space with the property that if $x$ is any point in it and $G$ any neighborhood of $x$, then $G$ contains a connected neighborhood of $x$. This is evidently equivalent to the condition that each point of the space have an open base whose sets are all connected subspaces. Locally connected spaces are quite abundant, for, as we have seen in Problem 32-5, every Banach space is locally connected.
We know that local compactness is implied by compactness. Local connectedness, however, neither implies, nor is implied by, connectedness. The union of two disjoint open intervals on the real line is a simple example of a space which is locally connected but not connected. A space can also be connected without being locally connected, as the following example shows. Let $X$ be the subspace of the Euclidean plane defined by $X=A \cup B$, where $A={(x, y): x=0$ and $-1 \leq y \leq 1}$ and $B={(x, y): 0<x \leq 1$ and $y=\sin (1 / x)}$ (see Fig. 26). $B$ is the image of the interval $(0,1]$ under the continuous mapping $f$ defined by
$$f(x)=(x, \sin (1 / x)),$$
so $B$ is connected by Theorem 31-B; and since $X=\bar{B}, X$ is connected by Theorem 32-B. $\quad X$ is not locally eonnected, however, for it is reasonably easy to see that each point $x$ in $A$ has a neighborhood which does not contain any connected neighborhood of $x$.

拓扑学代考

数学代写|拓扑学代写Topology代考|LOCALLY CONNECTED SPACES

$$f(x)=(x, \sin (1 / x))$$

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