I need a "mini-crashcours" concerning the space of germs of continuous functions in order to solve an exercise which requires me to show that limits in this space aren't always unique.
We have introduced these ad-hoc to illustrate nonhausdorff spaces, so I dont have any intuition about germs and am rather frustrated because I don't have a clear picture about them in my head.

In our functional analysis course we have defined the set of germs of continuous functions to be the $S/\sim$, where $$S:=\{ u:U\rightarrow \mathbb{R}:A\subseteq U\subseteq\mathbb{R}^n\ \text{open},u\big|_{U}\ \text{continuous}\}$$ and $A\subseteq\mathbb{R}^n$ is some fixed, closed set; $\sim$ is the equivalence relation defined by $$u:U\rightarrow \mathbb{R} \sim v:V\rightarrow \mathbb{R} \Longleftrightarrow \exists W\subseteq\mathbb{R}^n:A\subseteq W, W \ \text{open}, \ W\subseteq(U\cap V),u\big|_W =v\big|_W.$$This set is endowed with the topology "generated by the final structure with respect to the collection of maps $(F_U)_U:C(U,\mathbb{R})\rightarrow S/\sim,\ f\mapsto [f]$."

Concerning the things in the "...":
1) In which set do the $U$ from $(F_U)_U$ range ? The set of all open sets in $\mathbb{R}^n$ ?
2)With what topology is $C(U,\mathbb{R})$ endowed ? (For personal reasons asking the professor is unfortunately not an option) The topology of pointwise convergence, i.e. the product topology on $\prod_U \mathbb{R}$, restricted to the continuous functions ?

The Exercise I have to solve: If $n=1$ and $A=\{0\}$ I have to show the limits aren't unique.

Progress so far: Assuming that the answer to the questions above is "yes", the constant sequence of the equivalence classes of the constant $0$ function on some fixed open $U$, $[v]$, obviously converges in $S/\sim$ to $[v]$. But I have a hint that for continuous $u$ with $u(0)=0$ and $g_n(x):=u(x)h(nx)$ (where $h(x):=1$ for $|x|\geq2$ and $h(x)=0$ for $|x|\leq 1$ for $n\in \mathbb{N}$), we also have that the sequence $([g_n])_n$ converges to $[v]$.
But how do I prove the convergence of $([g_n])_n$?
(Please note, that I have to prove this "elementary", since I don't have any theorems at my disposal that tell me what kind of convergence a final structure induces...)

  • 3
    $\begingroup$ +1 for the well-formulated question and the title... $\endgroup$ Apr 29, 2014 at 8:32

1 Answer 1


In which set do the $U$ from (F_U)_U range ? The set of all open sets in $\mathbb{R}^n$ ?

Not quite. If $F_U(f) \in S/\sim$, then $f$ must be in $S$. That means $U$ must be an open set containing $A$.

With what topology is $C(U, \mathbb{R})$ endowed ? The topology of pointwise convergence, i.e. the product topology on $\prod_U \mathbb{R}$, restricted to the continuous functions ?

Generally, I would assume the compact-open topology. In this case it is enough to assume that the topology is uniform convergence or any coarser topology.

But how do I prove the convergence of $([g_n])_n$?

The only thing you need to know about the final topology, is that it makes every $F_U$ continuous. If you show that $g_n \to u$ in $C(U, \mathbb{R})$, it follows immediately from that continuity that $[g_n] = F_U(g_n) \to F_U(u) = [u]$ in $S/\sim$.

Edit: "Not Hausdorff" is a great understatement of how coarse the topology actually is. The fact that a constant sequence can converge to a limit unequal to the terms already proves that it is not T1, since $[u] \in \overline{\{[v]\}}$.

We can go further than that by taking two arbitrary continuous functions $v$ and $u$ such that $v(0) = u(0)$ and defining $g_n(x) = v(x) + (u(x) - v(x))h_n(x)$. By the same reasoning as before, we get $[u] \in \overline{\{[v]\}}$. Since we made no assumptions on $u$ and $v$, other than that they agree at 0, we can conclude that for any $c \in \mathbb{R}$ the subspace $\{ [f] \in S/\sim \,\mid f(0) = c \}$ is indiscrete.

I have not checked this formally, but from there I would guess that $S/\sim$ is homeomorphic to the product of $\mathbb{R}$ and an indiscrete space. That being said, I suspect that the objective of the exercise is precisely to demonstrate a way of proving things about a final topology without having an explicit description in terms of open sets, as such a description can be very complicated in more practical situations.

  • $\begingroup$ Before I accept your answer, could you please elaborate, why "the only thing that I need to about about the final topology,is that it makes every $F_U$ continuous" ? To me it seems that only knowing this rather obscures the fact, that this topology isn't hausdorff (this is just one property the description via this "universal property" seems to obscure - I could list many more). Whereas if I knew explicitly how the topology looks like, I could immediately see which neighborhoods of points are "too big" - and therefore prevent these points being a unique limit. $\endgroup$
    – user10324
    May 5, 2014 at 9:12
  • $\begingroup$ Thanks. So would it be correct to say, that using the final structure vs. directly the topology is something like a inclining balance between intuition and easy arguments towards the "easy arguments" side ? Because a description of the topology is very technical, if we extract it from the final structure and therefore the arguments get very technical, we accept to loose the intuition given directly by the topological and accept to work more "symbolically/algebraically" were the arguments are easier. $\endgroup$
    – user10324
    May 5, 2014 at 20:50
  • $\begingroup$ You could say that. I would also say that when we learn new ways of looking at things, intuition often needs some time to catch up. What seems like cheap trickery at first, may appear to be quite elegant and natural a year or so later. $\endgroup$ May 5, 2014 at 22:00

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