The Lie group $GL(n)$ being a manifold is locally pathconnected. Consider its connected component of the identity $C\subseteq\mathbb{R}^{n^2}$. What is a good way of showing that $C$ is a tame subset of $\mathbb{R}^{n^2}$ i.e., if for every $\epsilon>0$ there exists a $\delta>0$ such that if the Euclidean norm $xy<\delta$ for $x,y\in C$ then there is a path of length $<\epsilon$ in $C$ joining $x$ and $y$? An earlier version of the question used the term "locally pathconnected as a subset" which some users found confusing. Note that local pathconnectedness of subsets of Euclidean space is a completely standard usage. In particular the "comb" in the plane is not locally path connected in the plane precisely in this sense; see wiki.
The proposed fact is true, this is how I'd prove it.
The set $M_n(\mathbb R)\setminus\{0\}$ retracts on the unit sphere $S$ by linear rescaling $M\mapsto \frac{M}{M}$. Let $\hat C$ be the image of $C$.
Now $\Delta:=\{\det M=0\}$ is an algebraic hypersurface in $S$ (because the determinant is homogeneous) and $\hat C$ is a connected component of $S\setminus\Delta$. Let $d$ be the Euclidean distance induced on $S$.
Assume that $C$ is not tame. This gives a sequence $(M_n,N_n)_n$ of matrices in $S$ such that $d(M_n,N_n)<\frac{1}{n}$ and $M_n$ and $N_n$ cannot be joined by a path of length less than some $\epsilon$. It is so because what happens along halflines $\mathbb R_{>0} M$ is tame. Up to extract a subsequence we can form the limit $M\in S$.
 Clearly $M\in\Delta$ for otherwise one can take a ball of small radius $r>0$ in $\hat C$ around $M$, and all points in this ball are linked by a path of length at most $Kr$ for some universal constant $K=K_n$.
 If $M$ belongs to a regular point of $\Delta$ then locally you can find an analytic rectification of $\Delta$ onto a coordinate axis in $\mathbb R^{n^21}$. Here again a small halfball will give a contradiction (the fact that the rectification is sufficiently smooth implies that the distance is not too distorted). The point of the argument here is that you cannot "reach the other side" (even by taking a long detour) without crossing $\Delta$ itself.
 If $M$ belongs to the singular set of $\Delta$, you apply the same argument after desingularization of the hypersurface. According to this post the projection on the second factor $$R~:~\mathbb P_{n1}(\mathbb R)\times S\longrightarrow S $$ is a resolution of singularities when restricted to the set $X:=\{([v],M)~:~Mv=0\}$. Indeed $R(X)=\Delta$ and $X$ is smooth, for the differential of the map $$ \begin{eqnarray} F~:~\mathbb R^n\times M_n(\mathbb R) \longrightarrow & \mathbb R^n \\ (v,M)\longmapsto & Mv \end{eqnarray} $$ is $(h,A)\mapsto Mh+Av$, which has maximal rank $n$ if $v\neq0$ (take $h:=0$).


$\begingroup$ Cylindrical algebraic decomposition. I was just trying to determine more explicitly what sort of algebraicgeometric background goes into your proposed solution. $\endgroup$ – Mikhail Katz Apr 20 '16 at 12:39

$\begingroup$ It seems to me that your quotient of $GL_n$ is not compact (the function $tr^2/det$ is not bounded) and that instead something like the set of $n$tuples of linearly independent unit vectors needs to be used in the argument. $\endgroup$ – user83633 Apr 20 '16 at 13:14

$\begingroup$ I was just talking about the union of the two sets $A$ and $B$ where $A=\{(x,y,z): x^2+y^2\leq 1; z=0\}$ and $B=\{(x,y,z): x^2+z^2\leq 1; y=0\}$. $\endgroup$ – Mikhail Katz Apr 20 '16 at 13:30

$\begingroup$ @user83633: yes, obviously my answer was not wellwritten. I simplified it so that everything happens in ambient space. $\endgroup$ – Loïc Teyssier Apr 20 '16 at 13:37
In the paper
 Edward Bierstone: Differentiable functions, BOLL. SOC. BRAS. MAT.,VOL 11 N 2 (1980), 139190
one finds the following theorem, which states that for subanalytic subsets (those given by equations and $\le$inequalities involving real analytic functions) with dense interior the interior distance is locally Hoelder continuous with respect to Euclidean distance.
Theorem 6.17. Let $V$ be an open subset of $\mathbb R^n$, and $A$ a closed subanalytic subset of $V$ such that $Int(A)$ is dense in $A$. For every compact subset $L$ of $A$, there exists $c > 0$ and an integer $\alpha \ge 1$ such that any two points $b, y \in L$ can be joined by a semianalytic arc $\sigma$ in $A$ such that:
(1) $\sigma < cb  y^{1/\alpha}$
(2) $\sigma$ intersects $\partial A$ in at most finitely many points.

1$\begingroup$ Thanks. Could you possibly comment on how this takes care of the examples I mentioned above, such as the union of two disks in 3space, and the figure8 in the plane. I guess one has to look up the definition of "subanalytic"? $\endgroup$ – Mikhail Katz Apr 21 '16 at 10:19
This applies to the earlier version of the question:
It is an open subset since the determinant function is continuous, and open subsets are locally pathconnected.

$\begingroup$ Consider the complement $C$ in $\mathbb{R}^2$ of the closed interval $[1,1]$ on the $x$axis. Is $C$ locally path connected as a subset of the plane? $\endgroup$ – Mikhail Katz Apr 20 '16 at 11:36

1$\begingroup$ @user83633, I gave a definition of what I mean by local pathconnectedness as a subset in my question. Do me a favor and delete your answer. $\endgroup$ – Mikhail Katz Apr 20 '16 at 11:45

1$\begingroup$ No, please first find yourself a reasonable name for the concept you mean, and include a precise definition in the question. Then I will delete. $\endgroup$ – user83633 Apr 20 '16 at 11:48

1$\begingroup$ I did. What's unreasonable about "locally pathconnected as a subset"? $\endgroup$ – Mikhail Katz Apr 20 '16 at 11:50

3$\begingroup$ The name "locally pathconnected" is already taken. Maybe something like "uniformly locally pathconnected" would be better. $\endgroup$ – user83633 Apr 20 '16 at 11:51
One way is to reduce first to $x,y$ with $n$ distinct complex eigenvalues. Then the $n$ eigenvalues and eigenspaces of x are close to those of y. Now connect all those with paths (just make sure not to run through eigenvalue zero, and to keep invariance under complex conjugation).

$\begingroup$ Since the matrices are not assumed diagonalisable there may be no decomposition into eigenspaces. $\endgroup$ – Mikhail Katz Apr 20 '16 at 12:12

$\begingroup$ But there is an arbitrarily short path to a diagonalizable matrix. $\endgroup$ – user83633 Apr 20 '16 at 12:13

$\begingroup$ The problem is that things like Jordan block decomposition are not "stable", i.e., they tend to be discontinuous in the initial data. If you want to prove that all pairs of points can be so joined, this is an important concern. $\endgroup$ – Mikhail Katz Apr 20 '16 at 12:15

$\begingroup$ The characteristic polynomial depends continuously on the matrix, the set of zeros of a polynomial depends continuously on the coefficients, and the the function sending a rank n1 matrix to its kernel is also continuous. Using all this, the set of eigenvalues and eigenspaces of two close diagonalizable matrices are close. $\endgroup$ – user83633 Apr 20 '16 at 12:20

1$\begingroup$ You are right. Moving all eigenspaces of x to eigenspaces of y simultaneously is a problem. $\endgroup$ – user83633 Apr 20 '16 at 12:27