connected space কী?

Example 4.28

$GL_n(\R)$ is not connected: $\det$ is continuous (a polynomial in the entries) onto $\R^*$, which is not connected; the preimages of $\R_+^*$ and $\R_-^*$ split $GL_n(\R)$. (Each piece is in fact path-connected — a pleasant exercise beyond our needs.) By contrast $GL_n(\C)$ is path-connected: Exercise 4.10.

Example 4.29 (A fixed point from connectedness alone)

Every continuous $f \colon \intcc01 \to \intcc01$ has a fixed point — no contraction hypothesis, no iteration. Consider $g(x) = f(x) - x$, continuous on the connected $\intcc01$:

and the intermediate value theorem (Theorem 4.27 (2)) delivers a zero of $g$, i.e. a fixed point of $f$. Contrast with Banach (Theorem 4.12): here existence is topological and free, but uniqueness and the algorithm are lost — $f = \mathrm{id}$ has every point fixed, and iteration of a non-contracting $f$ may cycle forever. The two fixed point theorems of this chapter answer different questions with different currencies.

Example 4.30 ($\R$ and $\R^2$ are not homeomorphic)

Connectedness is a topological fingerprint. Suppose $h \colon \R^2 \to \R$ were a homeomorphism (a continuous bijection with continuous inverse). Remove one point $a \in \R^2$: the restriction $h \colon \R^2\setminus\{a\} \to \R\setminus\{h(a)\}$ is still a homeomorphism. But $\R^2$ minus a point is path-connected — join any two points by a segment, detouring along a second segment through an auxiliary point if $a$ blocks the direct one — hence connected (Theorem 4.27 (3)); while $\R$ minus a point splits into two nonempty open half-lines: not connected. Connectedness is preserved by continuous maps: contradiction. The plane and the line are genuinely different as topological spaces — a fact that cardinality alone (Exercise 1.3-style bijections do exist!) is too coarse to see.

Example 6.22 (The topologist’s sine curve)

Let $\Gamma = \{(x, \sin\frac1x) : 0 < x \leq 1\}$ and $S = \bar\Gamma = \Gamma \cup (\{0\}\times[-1,1])$ (every point $(0, y)$, $\abs y \leq 1$, is a limit of points of $\Gamma$: solve $\sin\frac1x = y$ near $0$). Then $S$ is connected — closure of the connected $\Gamma$, a continuous image of $(0, 1]$ (Theorem 6.20(3)) — but not path-connected: a path from $(1, \sin 1)$ to $(0,0)$ would have to traverse abscissas $\to 0$ while the ordinate oscillates between $\pm1$; Exercise 6.9 makes this rigorous. Connectedness and path-connectedness genuinely differ.