Are pictures legitimate as a proof in mathematics? While I'm studying Topology (teaching it myself with videos and books) I've seen some 'proofs' with pictorial approach and solution, I haven't seen it before. So is it legitimate?
 A: How about this one? It's the simplest (and only) one I can think up just like that, that has to do with topology in that it is about continuity:
Convex functions $f \colon \mathbb{R} \rightarrow \mathbb{R}$ are continuous:

The interpretation being: The essential property of a convex function is that all line segments between points on the graph lie entirely above the graph. Assume you have a convex, discontinuous function, as shown - then there will be a line segment connecting two points on the graph, which passes under a portion of the graph.
Edit:
Math asks in the comments what it is exactly that this argument proves, so here's my attempt at clarifying:
There are two things to consider here, really: 
The concept of continuity: is the rough graph really representative of what it means for a function to be continuous, or in this case discontinuous? The definition of continuity goes something like:
Given a function $f \colon A \rightarrow B$, with $A, B \subseteq \mathbb{R}$. Pick a point $x \in A$; then $f$ is said to be continuous in $x$, if $\forall \epsilon > 0, \exists \ \delta > 0 \colon |x - x_0| < \delta \Rightarrow |f(x) - f(x_0)|$ - the epsilon-delta argument feared and hated by young maths student everywhere. But what does it actually mean? What it says, really, is that no matter how small a 'box' you choose around $(x,f(x))$, you will always be able to find a bit of the function's graph near the point $(x,f(x))$. Dicontinuous at $x$, in contrast, means that you can find a value of $\epsilon$ where this is no longer the case - there will be a hole, where the graph jumps up or down, as illustrated in my rough drawing.
The other thing to keep in mind is the definition of convexity: A convex graph is one that always 'bends upward' - or in other words, any straight line between two points on the graph will always lie above the graph. 
So, what did the argument above prove? Well, the drawing shows a 'convex' graph with a hole in it: 


*

*I start by assuming that there is such a thing: a convex function that isn't continuous everywhere, and hopefully my considerations about what discontinuous means, have convinced you that this drawing is a valid representation of all such graphs. 

*I then draw a line between two points on that graph, which as you can see is not entirely above the graph. I only need to find one such line to prove that this graph isn't convex, since the definition says that there mustn't be any lines that go below the graph.


This shows that a discontinuous function cannot be convex: it violates the definition. In other words, if a function is convex, then it can't be discontinuous - so it must be continuous.
This type of proof goes under the name reductio ad absurdum, and it is quite possibly one of the most annoying tools in logic: it so often leads to proofs where you know something is true, but you have no way of constructing a good example. Well, that's my opinion, any way.
