Suppose $X \subset \mathbb{R}^n$ is connected. Let $f:X\longrightarrow \mathbb{R}^n$ be a continuous function. Prove that if $f$ is constant on the path-connected components of $X$, then $f$ is constant on $X$. I was able to prove this in the case where $X$ has a finite number of path-connected components. In fact, we can proceed by induction. Starting with a path-connected component, its closure must intersect the closure of another path-connected component; by continuity the constant on these must agree. I have no idea how to handle the general case, any hint is appreciated. The case where $f$ is uniformly continuous (or Holder) is of interest too. In that case, we can reduce the problem to a compact $X$, if that is of help. I wonder if the cardinality of path-connected components of a (compact) connected subset of $\mathbb{R}^n$ must be countable. If that was the case, maybe an induction could be useful. Thanks!
Real Analysis – Constant on Path-Connected Components Implies Constant
analysiscontinuitygeneral-topologymetric-spacesreal-analysis
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Another way would be to prove that every elementary matrix $E$ is connected in $GL_n(\mathbb{R})$ to either $I_n$ or $\begin{pmatrix} -1 & 0 \\ 0 & I_{n-1} \end{pmatrix}$. Using Gaussian elimination, every $n \times n$ real matrix $A$ can written in the form $A = E_1 E_2 \cdots E_n R$ where $R$ is the row-reduced echelon form of $A$ and the $E_i$ are elementary matrices. If $A$ is invertible, then $R = I_n$, so this just says every $A \in GL_n(\mathbb{R})$ can be expressed as a product of elementary matrices: $$A = E_1 E_2 \cdots E_n.$$ Continuously deforming each $E_i$ into either $I_n$ or $\begin{pmatrix} -1 & 0 \\ 0 & I_{n-1} \end{pmatrix}$ as appropriate shows that $A$ can also be continuously deformed to either $I_n$ or $\begin{pmatrix} -1 & 0 \\ 0 & I_{n-1} \end{pmatrix}$, depending on whether an even or odd number of the latter matrix occurs in the product when the deformation is completed. This implies $GL_n(\mathbb{R})$ has at most two path-components, and you have already observed there are at least two, using the determinant.
First, argue that every elementary matrix $E$ can be connected either to $I_n$ or to $I_n$ with a single diagonal entry replaced by $-1$. There are three types of elementary matrix, hence three cases to consider.
- $E$ is $I_n$ plus a single off-diagonal entry $\lambda$. In this case, $E$ is connected to $I_n$ by a path of elementary matrices of the same type. Just continuously decay $\lambda$ to zero.
- $E$ is $I_n$ with a single diagonal entry replaced by a nonzero scalar $\lambda$.
- If $\lambda < 0$, then $E$ is connected to $I_n$ by a path of elementary matrices of the same type. Just continuously change $\lambda$ to $1$ without crossing $0$.
- If $\lambda < 0$, then $E$ is connected to $I_n$ with a single diagonal entry replaced by $-1$ by a path of elementary matrices of the same type. Just continuously change $\lambda$ to $-1$ without crossing $0$.
- $E$ is a transposition. For example, $E = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\\ \end{pmatrix}$. Note $E = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\\ \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{pmatrix}$ and the matrix on the right is connected to $I_n$ via $\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos(t) & -\sin(t) \\ 0 & \sin(t) & \cos(t) \\ \end{pmatrix}$ so that $E$ is connected to $\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\\ \end{pmatrix}$. In general, when $E$ is any transposition, it is connected to $I_n$ with a single diagonal entry replaced by $-1$.
The argument will be complete provided that any matrix $B$ which is $I_n$ with a single diagonal entry replaced by $-1$ is connected to the particular example $\begin{pmatrix} -1 & 0 \\ 0 & I_{n-1} \end{pmatrix}$. Indeed, if one has, say, $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ then one can use the path $\begin{pmatrix} \cos(t) & 0 & -\sin(t) \\ 0 & 1 & 0 \\ \sin(t) & 0 & \cos(t)\end{pmatrix} B \begin{pmatrix} \cos(t) & 0 & -\sin(t) \\ 0 & 1 & 0 \\ \sin(t) & 0 & \cos(t)\end{pmatrix}^{-1}$ to continuously shift the $-1$ to the upper-left corner.
If you have Munkres (2nd ed.): Thm 25.4 states:
A space $X$ is locally path-connected iff for every open set $U$ of $X$, each path-component of $U$ is open in $X$.
So if indeed $U$ is open and connected it has (as all spaces) a decomposition into path-components, which are open in $X$ and thus open in $U$ too, and by being a partition, they are also closed (the complement is also a union of open sets) in $U$. So by connectedness there can be only one path-component.
25.5 even says (part 2 of it)
If $X$ is a topological space, and $X$ is locally path-connected, its components and path-components coincide.
Apply this to $X=U$ (which is locally path-connected as an open subspace of a locally path-connected space) and you're done right away.
Note, this is assuming you use the same definition of local path-connectedness as Munkres does (which is non-standard): every neighbourhood $U$ of $x$ contains a path-connected neighbourhood $V$ of $x$.
The definition in e.g. Engelking is:
for every open set $U$ and every $x \in U$ there is an open neighbourhood $V$ of $x$ such that for any $y \in V$ there is a path $p: [0,1] \to U$ connecting $x$ to $y$.
Note that $V$ is not supposed to be itself path-connected, as Munkres does. So the latter has a stronger notion, so maybe this fact only holds for the stronger notion; at least the proof does.
Best Answer
One example where the claim that you are trying to prove is false is given by the pseudo-arc $A$, which is a nondegenerate planar continuum (i.e. a compact connected subset of the plane containing more than one point). The pseudo-arc has a number of remarkable properties, in particular:
(a) It is homeomorphic to any of its nondegenerate subcontinua.
(b) it is indecomposable, i.e. it cannot be represented as a union of two proper subcontinua.
From these two properties, it is easy to conclude that every continuous map $[0,1]\to A$ has to be constant. In particular, every path-connected subset of $A$ is a singleton. It is also clear that $A$ admits a nonconstant continuous function, e.g. the restriction of a coordinate function.
Edit. For the sake of completeness, here is a proof (one of many) that every continuous map $f: I=[0,1]\to A$ is constant. Suppose that there exists a nonconstant map $f$. Pick two distinct points $x, y\in f(I)$; (after renaming $x, y$ if necessary) there exists a subinterval $[a,b]\subset [0,1]$ such that $f(a)=x, f(b)=y$ and $f^{-1}(\{x,y\})\cap [a,b]=\{a,b\}$. The image $f([a,b])$ is connected, compact, hence, is a nondegenerate continuum, hence, is homeomorphic to $A$. Thus, by restricting $f$ to $[a,b]$, we reduce to the case $[a,b]=[0,1]$. Since $f^{-1}(\{x,y\})=\{0,1\}$, the subcontinua $f([0, 0.5])$ and $f([0.5, 1])$ in $A$ are both proper, nondegenerate and their union is $A$. This contradicts the fact that $A$ is indecomposable.