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General Topology: Chapters 1-4

Sets

Set Closure

Statements

Theorem. Let $(X,d)$ be a metric space and $S\subseteq X$. If $\mathrm{cl}(S)$ is the metric set closure, then it also is the topological set closure.

Proof. Follows from the topological space induced by the metric space.

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Set Boundary

Statements

Definition. (@Bou95, Definition 11). Let $(X,\tau )$ be a topological space, and $S\subset X$. A point $x\in X$ is said to a boundary point of $S$ if $x$ belongs to the closure of $S$ and its complement, ie, $x\in \mathrm{cl}(S)\cap \mathrm{cl}(X\setminus S)$. The set of all boundary points is said to be the boundary of $S$.

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Functions

Statements

Definition (@Tu11 pp. 327, @Bou95, Chapter 2, Definition 1). Let $(X,{\tau }_X)$ and $(Y,{\tau }_Y)$ be topological spaces. A mapping $f:X\to Y$ is said to be continuous at a point $x\in X$ if, for every topological neighborhood $N_Y$ of $f(x)\subset Y$, there exists a topological neighborhood $N_X$ of $x\in X$ such that the relation $x\in N_X$ implies $f(x)\in N_Y$. In addition, $f$ is continuous on $X$ if it is continuous at every point of $X$.

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