What is a topology? Having read through the mathematical definition of endowing a set with a topology I must admit that I'm still struggling to conceptualise what such a mathematical construct is. 
I've read articles that talk about a topology on a set as defining a notion of "nearness" between elements of that set without needing to introduce the notion of distance (a metric). Is what is meant by this that, say for example I have a set $X=\lbrace a,b,c,d\rbrace$, then I can define a topology on this set $\tau = \lbrace \emptyset ,\lbrace a\rbrace, \lbrace a,b\rbrace, \lbrace a,b,c\rbrace, \lbrace a,b,c,d\rbrace\rbrace$. Now, as $\lbrace a\rbrace\cup\lbrace a,b\rbrace = \lbrace a,b\rbrace$ and $\lbrace a\rbrace\cup\lbrace a,b\rbrace\cup\lbrace a,b,c\rbrace = \lbrace a,b,c\rbrace$, would it be correct to say that $a$ is "nearer" to $b$ than it is to $c$ as $b$ is contained within a smaller neighbourhood of $a$? Is this what is also meant when people speak of two objects being topologically equivalent, as although they may look very different, the relationships between the points remain the same, i.e. $b$ is still "nearer" to $a$ than $c$ even though there may be a huge number of different points added to the neighbourhoods of each of them under some continuous deformation?
 A: A topology doesn't encode the idea of 'nearness' between points in the same way a metric does.  Instead, it tells us when a point is 'near' to a set.  
If we restrict ourselves to Hausdorff spaces for simplicity, then two points are either the same, or they are different points, and there is no topological distinction between two points at a distance of $1$ and two points at a distance of $2$.  However, it does make sense to say that the point $0$ is 'near' to the open interval $(0,1)$ - it is not contained in that interval, but there are points in $(0,1)$ that get arbitrarily close to $0$.  
If $X$ is a topological space, $x\in X$ and $S\subset X$ is any subset, we say $x$ is near to $S$ if $x$ is contained in the closure of $S$ - the intersection of all closed sets containing $S$.  For example, the closure of $(0,1)$ in $\mathbb R$ is $[0,1]$, which contains $0$, so we can say that $0$ is near to $(0,1)$.
$\DeclareMathOperator{\cl}{cl}$
We even get an alternative definition of a topology this way.  Given a set $X$, a closure operator on $X$ is a function $\cl\colon\mathcal P(X)\to\mathcal P(X)$ such that:


*

*$\cl(\emptyset)=\emptyset$

*For all $A\subset X$, $A\subseteq \cl(A)$

*For all $A,B\subset X$, $\cl(A\cup B)=\cl(A)\cup\cl(B)$

*For all $A\subset X$, $\cl(\cl(A))=\cl(A)$


Think of $\cl$ as being the function that takes a subset $A\subset X$ and returns the set of points of $X$ that are near to $A$.  Then the axioms can be stated as:


*

*No point is near to the empty set.

*Every point of $A$ is near to $A$.

*The points of $X$ that are near to $A\cup B$ are the points that are near either to $A$ or to $B$.

*If $x$ is near to the set of points that are near to $A$, then $x$ is near to $A$.


Given a closure operator $\cl$, we can define a topology on $X$ by setting the closed sets to be those sets $A\subset X$ such that $\cl(A)=A$.  In particular, by axiom (4), $\cl(A)$ is always a closed set for any $A\subset X$.  
Exercise: check that the family $\mathcal K$ of subsets $A\subset X$ satisfying $\cl(A)=A$ is closed under intersections and finite unions; i.e., it is the family of closed sets for a topology on $X$.
Conversely, given a topological space $X$, we can define $\cl$ to be the usual closure operator on $X$; i.e., 
$$
\cl(A)=(\textrm{intersection of all closed sets containing }A)
$$
Exercise: Check that $\cl$ satisfies the axioms of a closure operator given above.  
The third exercise should convince you that the definition by closures gives us an alternative definition of a topological space:
Exercise: Let $(X,\tau)$ be a topological space, and let $\cl$ be the usual closure operator as defined immediately above.  Show that the family $\mathcal K$ induced by $\cl$ is precisely the family of closed sets of $X$.
Conversely, let $(X,\tilde{\cl})$ be a set together with a closure operator $\tilde{\cl}$, and let $\mathcal K$ be the induced family of closed sets.  Define a topology $\tau$ on $X$ by $\tau = \{X\setminus L\colon L\in\mathcal K\}$.  Show that the usual closure operator on $(X,\tau)$ coincides with $\tilde{\cl}$.  
Extension: Recall that a function $f\colon X\to Y$ between topological spaces $(X,\tau),(Y,\sigma)$ is called continuous if $f^{-1}(U)$ is open in $X$ whenever $U$ is open in $Y$ (if you like, $f^{-1}(U)\in\tau$ for all $U\in\sigma$).  Give a definition of a continuous map in terms of closure operators.  
Further extension: Study what happens if we omit axiom (4) in the definition of a closure operator (and have a chat with me to discuss what you're discovered!)
