# Classify all alphabets into homeomorphism classes $\{ M N B H\}$,

classify all alphabets into homeomorphism classes $\{ M N B H\}$, what does it mean by homeomorphism classes? They're looking upon letters as drawings of topological spaces, perhaps!?

connectedness distinguishes $B$ from other $3$ as if we remove mid point of vertical line of $B$, it is still one single component, but if we remove one point from other three they will be into two component. but if we remove 2 point suitably from $H$ it will be broken into $5$ component, but the case is not with $M$ and $N$ they will be broken into 3 component, so the homeomorphism classes?

$\{H\},\{M,N\},\{B\}$, am I right? is there any other approach?

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Yes, I think you got it right. Except at the end you reversed 2 and 3. – user31373 Aug 14 '12 at 22:34
if I remove left and right horizontal point of middle line of $H$ it leaves $5$ component sir. – Un Chien Andalou Aug 14 '12 at 22:37
OK. But it's probably easier to consider what happens when just one point is removed. You sometimes get 3 pieces from H, but never from M or N. – user31373 Aug 14 '12 at 22:39
[looks closely at the letters] wait a moment - are we talking about serif or sans-serif font? – user31373 Aug 14 '12 at 22:40
Question of NBHM 2010 Topology 3.10. – Dutta Nov 7 '13 at 14:23

I'll describe a general approach here, using for illustration purposes the serif font that appears in the question. We are given compact connected Hausdorff spaces, also known as continua. A point $x$ of continuum $C$ is a cut point if $C\setminus\{x\}$ is not connected. The number of components of $C\setminus\{x\}$ is the order of cut point $x$. The cardinality of the set of cut points of order $n$ is a topological invariant of the space.
All of our continua have infinitely many cut points of order 2, so that does not help much (although we could look at the structure of the set of all such cut points, e.g., the number of its components). Let's count the cut points of order 3: $M$ has 4, $N$ has $3$, $B$ has none and $H$ has $6$. Hence, none of them are homeomorphic in the serif version.
In addition, $B$ is distinguished from the rest by having infinitely many non-cut points. It's a very neat theorem of general topology that every nondegenerate continuum has at least two non-cut points.