Topology as a generalization of the open sets of the reals

As far as I can make out, topology is a generalization of the properties of open sets of the reals; this is evident in the terminology; for a set $X$, subsets of this space are actually called open if they are a member of the topology on that set. Topology extracts the "main properties" of open sets by requiring that arbitrary unions and finite intersections of open sets are open.

I've also gathered that topology is useful because it creates a setting in which we can do analysis on a space, as it gives us an idea of convergence, continuity of maps etc. What I don't understand is why this is the case. Why do we care about open sets? Is real analysis based around open sets? The definition of the continuity of a function $f:\mathbf{R} \rightarrow \mathbf{R}$ at a point $c\in \mathbf{R}$ is

$\forall \epsilon >0$ $\exists \delta >0$ such that $\forall x\in (c-\delta , c+\delta)$ we have $f(x)\in (f(c)-\epsilon, f(c)+\epsilon )$,

but we could replace the open sets here with closed sets and have the same definition. In topology a continuous function $f:X\rightarrow Y$ between topological spaces $(X, \tau_X)$, $(Y,\tau_Y)$ is defined to be one such that for any open set $V\in \tau_Y$, its pre-image $f^{-1}(V)$ is open in $\tau_X$. How does this relate to the definition of a continuous function in the setting of real analysis? Thanks for any replies!

• Also, a continuous map is one such that $f^{-1}(F)$ is closed for every closed $F\subset Y$. Since you took the "continuous at a point" viewpoint for real functions, take it also in general, and note that the important thing there are neighbourhoods. $f$ is continuous at $c$ if and only if $f^{-1}(N)$ is a neighbourhood of $c$ for every neighbourhood $N$ of $f(c)$. – Daniel Fischer Oct 8 '14 at 17:24

You're slightly mistaken about "we could have replaced this with closed sets". In choosing $\epsilon > 0$ and $\delta > 0$, we're essentially covering all open sets around $x$ and $f(x)$, in the sense that every such set contains one of these intervals. The corresponding statement for closed sets would be

$\forall \epsilon \ge 0$ $\exists \delta \ge 0$ such that $\forall x\in [c-\delta , c+\delta]$ we have $f(x)\in [f(c)-\epsilon, f(c)+\epsilon ]$,

which is true for ANY function: you just pick $\delta = 0$.

There really is something different about "less than or equal to" and "less than", and this definition of continuity happens to find just that sweet spot where it matters.

Open sets is a rather abstract approach to topology. I would prefer some other introduction to this topic, for example the concept of closure.

In an open interval $(a,b)$ there are points in the interval very near $a$ but $a$ is not included. Among those points that doesn't belongs to $(a,b)$, $a$ and $b$ are special, because they are very near the set.

A topology on a set $X$ is a way to determine such special points. As $x\in A$ is the condition for the points included in $A$, $\;x\in \bar A$ is the condition for the points in or near the set $A$. The set $\bar A$ is called the closure of $A$.

Any relation $\propto\,\subseteq X\times \mathcal P(X)$ that extends $'\!\!\!\in'$ in the way that,

1. $x\in A\Rightarrow x\propto A$
2. $\neg\exists x\in X:x\propto\emptyset$
3. $x\propto A \subseteq B \Rightarrow x\propto B$
4. $x\propto A\cup B\Rightarrow x\propto A \vee x\propto B$

defines a closure operator on subsets of $X$ by $x\in \bar A \Leftrightarrow x\propto A$. And since all the sets $\bar A$ satisfies the axioms for closed set and since every closed set is among the $\bar A$-sets and any open set is the complement of a closed set, the closure operator define $\tau$, the set of all open sets in $X$.

Open sets are efficient tools to prove important things in topology. The $\epsilon$-$\delta$ technique is based on the open sets these represent, but in general topology there is no real numbers so the corresponding technique is applied to open sets.

To prove that a function $f:X\rightarrow Y$ is continuous it is often best to prove that: $V$ open in $Y$ implies that $f^{-1}(V)$ is open i X. This is an abstract condition that also works in analysis, which can be replaced by an equivalent and more intuitive condition in topology (and analysis of course): given two topological spaces $X$ and $Y$, then a function $X\overset{f}\longrightarrow Y$ is continuous if and only if, $$M\subseteq Y\Rightarrow(x\in \overline{f^{-1}(M)}\Rightarrow f(x)\in \overline{M})$$