Show that the letters X and I (thought of as topological spaces) are not homeomorphic I am reading Hajime Sato's: Algebraic Topology, an Intuitive Approach.  His Sample Problem 1.3 is: Show that the topological spaces X and I are not homeomorphic.  (Note that this requires a font where I appears as a simple straight vertical line.) 
Sato argues by contradiction.  Brief sketch of his argument: Suppose there existed a homeomorphism $f: X \to I$.   Remove the point $x_0$ which lies at the crossing point of X.  Then the restriction of $f$ to the new domain is still homeomorphic.  But $X-x_0$ consists of four disjoint line segments (each half open), and $I-f(x_0)$ consists of two disjoint line segments (each half open).
Therefore, he concludes, the spaces aren't homeomorphic.  
I don't understand how the conclusion (in the final sentence) follows.   I know that two spaces are homeomorphic if there exists a bijective continuous map  between them with a continuous inverse.  And I know the characterization of continuity in terms of open sets.   But somehow I am not seeing it.   I feel like I'm missing something really obvious.  Any ideas?
 A: A homeomorphism is an isomorphism between topological spaces.  That is to say, it preserves the topological structure.  
Let's forget the definition of a homeomorphism in terms of continuous maps for now, and try and construct it ourselves.  If $X$ is a topological space, and $Y$ is a topological space that is 'the same' as $X$, up to renaming of the points, then that means that there is a bijective function
$$
f\colon X\to Y
$$
taking a point $x\in X$ to the corresponding point $y\in Y$.  As a silly example, the topological spaces $\mathbb R\times\{0\}$ and $\mathbb R\times\{1\}$ are homeomorphic.  The homeomorphism $f$ takes the point $(x,0)$ to the point $(x, 1)$.  
Now we want $f$ to preserve the topological structure somehow.  Since we want $X$ and $Y$ to be 'the same', up to the 'renaming' provided by the function $f$, then the open subsets of $Y$ should be precisely those sets $V\subset Y$ that correspond to open subsets $U\subset X$, via the map $f$.  Putting this into mathematical language, we see that:

$V\subset Y$ is open in $Y$ if and only if the set $\{f^{-1}(y)\;\colon\;y\in V\}\subset X$ is open in $X$.

(remembering that $f$ is a bijection, so it has an inverse $f^{-1}$).  If we think a bit harder, we can see that this is equivalent to the usual definition of a homeomorphism (left as exercise...)
This is very powerful.  It means that any statement we can make about $X$ using purely topological language must also be true for $Y$.  For example, if we say '$X$ has $4$ connected components', then that is really a statement about the open sets of $X$ - it is a topological statement.  If $X$ is homeomorphic to a space $Y$ then $Y$ must have $4$ connected components too.
On the other hand, if we say $X$ has diameter $2$, then that is not a topological statement, and we should not expect it to be preserved by homeomorphism.  For instance, the unit circle, which has diameter $2$, is homeomorphic to the unit square, which has diameter $\sqrt{2}$, and to a larger circle, of diameter $1000$.  
Of course, you should still prove that $(\text{number of connected components})$ is a homeomorphism invariant, but in general, you can be sure that any definition made purely in terms of open sets and relations between them and the points of the space will be preserved by homeomorphisms.   
A: If I is disconnected by removing the point f(xo) and X by removing the point of conjunction xo ... there are two connected components of I while 4 connected components of X ... so if f is homeomorphism images of connected components are also connected ... is it possible here ? Can you map the 2 connected components to connected components of X while f being a bijection at the same time ? 
A: The issue is that connectedness is preserved by homeomorphisms. More exactly, two homeomorphic spaces must have the same number of connected components. If $X \cong I$, then $X \setminus \{\rm crossing~point\} \cong I \setminus \{\rm one ~point\}$. But $X \setminus \{\rm crossing~point\}$ has four connected components, while $I \setminus \{\rm one ~point\}$ hasn't. Done.
