Approach to Defining a Topology

Here, I will be discussing my perspective on the motivation for Topology. The following information comes from my notes on Topology found here.


We are motivated by the following definitions and theorems:

  • Definition. (Open and Closed) A set G\subset\mathbb{R} is open if \forall{x}\in{G} \ \exists\epsilon>0 such that (x-\epsilon,x+\epsilon)\subset{G} (interior points). A set F\subset\mathbb{R} is closed if F^c=\mathbb{R}-F is open.
  • Ex) (a) Let [0,1]\subset\mathbb{R}. Then to be an open set, \forall{x}\in{G} \ \exists\epsilon>0 such that (x-\epsilon,x+\epsilon)\subset[0,1]. However, if we chose 0 or 1 for x, there is no (x-\epsilon,x+\epsilon) in [0,1]. Hence, [0,1] is not an open set. On the other hand, we define [0,1]^c=\mathbb{R}-[0,1], which is open. Therefore, by the definition of a closed set, [0,1] is a closed set. (b) Let (0,1)\subset\mathbb{R}. Then we take \forall{x}\in(0,1), \exists\epsilon>0 such that (x-\epsilon,x+\epsilon)\subset(0,1). We can see that all points x contained in (0,1) have a (x-\epsilon,x+\epsilon)-interval. I.e., for x=0.0001, \ (0.00001,0.001)\subset(0,1). Therefore, (0,1) is an open set.
  • Theorem. (Open Sets) (a) If \{G_{\lambda}:\lambda\in\Lambda\} is a collection of open sets, then \cup_{\lambda\in\Lambda} is open. (b) If \{G_k:1\leq{k}\leq{n}\} is a finite collection of open sets, then \cap^n_{k=1}G_k is open. (c) Both \emptyset and \mathbb{R} are open.
  • Theorem. (Closed Sets) (a) If \{F_{\lambda}:\lambda\in\Lambda\} is a collection of closed sets, then \cap_{\lambda\in\Lambda}F_{\lambda} is closed(b) If \{F_k:1\leq{k}\leq{n}\} is a finite collection of closed sets, then \cup_{k=1}^n{F_k} is closed(c) Both \emptyset and \mathbb{R} are closed.
  • Definition. (Limit Points) x_0 is a limit point of S\subset\mathbb{R} if \forall\epsilon>0, the (x_0-\epsilon,x_0+\epsilon)\cap{S}\backslash\{x_0\}\neq\emptyset.

From this, we can form our definition of a topology.

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