I am a beginner of AT and I cannot distinguish what we mean by 'connected', that is
connected
locally connected
path connected
could anyone help?
I am a beginner of AT and I cannot distinguish what we mean by 'connected', that is
connected
locally connected
path connected
could anyone help?
On
A space $X$ is connected if whenever it is decomposed as the union $A \cup B$, $A,B$ nonempty, then $\overline{A} \cap B \neq \varnothing$ and $\overline{B} \cap A \neq \varnothing$.
Example: The real line.
A space $X$ is locally connected if for each $x\in X$, and each neighborhood $U$ of $x$, there exists a connected neighborhood $V$ of $x$ that is contained within $U$.
An example of a space that is not locally connected is: $X = \{0\} \cup \{1/n : n = 1,2,...\}$ with the subspace topology from the real line.
Note that connected does not imply locally connected. A common nonexample is the comb space (connected but not locally connected):
http://en.wikipedia.org/wiki/Comb_space
A space $X$ is path-connected if any two points $x,y \in X$ can be connected by a path, where a path is a continuous function $\gamma: [0,1] \to X$. Note that a path-connected space IS connected.
However, a connected space is not always path connected. The textbook example of this is http://en.wikipedia.org/wiki/Topologist%27s_sine_curve
Try working through the definitions of each of the terms and then looking at some examples/counterexamples to build some intuition.
Connected: A space is connected if, for any $A$ and $B$ non empty open sets, then $X = A \cup B$ implies that $X = A$ or $X = B$. In other words, we can't break up $X$ into two disjoint open halves. Or, any nonempty closed and open set is the entire space $X$.
A basic idea here is that of a connected component of a space. Given a space $X$, a connected component is a maximal connected subset $A$. A connected space has exactly one connected component.
Path Connected: A space is path connected if given any two points $a$ and $b$ I can draw a path between them - that is, there is a continuous function $f: [0,1] \to X$ such that $f(0) = a$ and $f(1) = b$. That is, $X$ consists of exactly one path component.
This implies connected, but the converse is not true. The standard counter example is the topologists sine curve, which is connected but not path connected. (Google it.)
However, if your space X is locally path connected (in the sense that every point has a neighborhood that is path connected) and also connected, then X will be path connected - for now the path components are open, meaning that there cannot be more than one of them, since they are automatically disjoint. (If two path connected sets are not disjoint, I can connect any point in one to the other by stopping in the intersection first.) This tells us, for instance, that any open subset of $R^n$ that is connected is automatically path connected, since points in $R^n$ are locally path connected.
Locally + property P typically means that every point $x$ has a neighborhood $V$ so that the space $V$ has the property $V$. For example, you might say that the circle is locally euclidean, for it is locally like a piece of $R^1$.
Hope that helps some.