# The set $C[0,1]$ is nowhere dense in $D[0,1]$.

$$D[0,1]$$ be the space of real functions $$x$$ on $$[0,1]$$ that are right continuous and have left hand limit.

On the other hand $$C[0,1]$$ is set of all continuous real functions on that interval.

Considering the underlying topology is Skorohod topology. I try to show that the set $$C[0,1]$$ is nowhere dense in $$D[0,1]$$. How to do that?

One idea is,

Suppose consider the ball in $$D[0,1]$$ i.e. $$B(f,\epsilon) = \{g\in D[0,1] | d(f,g) < \epsilon\}$$, where $$d$$ is Skorohod metric. Now it is enough to show that $$C[0,1] \cap B(f,\epsilon) = \emptyset$$. Then we can conclude that $$C$$ is nowhere dense.

But how to show this claim?

• @AugustoSantos In $D$ uniform norm doesn't make sense that's why Skorohod metric required. I think. Commented Apr 27, 2022 at 14:47
• $C$ is a proper closed linear subset of $D$. Thus it has empty interior. Commented Apr 27, 2022 at 17:47
• @OliverDíaz how to show that $C$ is proper closed linear subset of $D$? Commented Apr 28, 2022 at 1:53
• The topology $d_0$ restricted to $C$ coincided with the uniform topology in $C$. Commented Apr 28, 2022 at 2:45

An explicit development.

Claim 1. Let $$g\in D\left[0,1\right]$$ be a càdlàg function with a jump of size $$J>0$$ at $$t\in \left(0,1\right)$$, i.e., $$J=\left|g(t)-\lim\limits_{x\rightarrow t-} g(x)\right|$$. Then, $$\|h-g\|\geq \frac{J}{2}\tag{\star}\label{star},$$ for any $$h\in\mathcal{C}\left[0,1\right]$$.

Proof. Remark that $$|h(t)-g(t)|+|h(t)-\lim\limits_{x\rightarrow t-} g(x)|\geq J/2$$. Consider $$\delta>0$$ small enough so that i) [h is continuous] $$|h(x)-h(t)|<\varepsilon/3$$ for all $$x\in\left(-\delta,\delta\right)$$; ii) [$$g$$ is right-continuous] $$|g(x)-g(t)|<\varepsilon/3$$ for all $$x\in\left(0,\delta\right)$$; and iii) [$$g$$ has left-limit] $$|g(x)-\lim\limits_{x\rightarrow t-} g(x)|<\varepsilon/3$$ for all $$x\in\left(-\delta,0\right)$$. Then, $$|g(x)-h(x)|\geq J/2-\varepsilon$$ for all $$x\in \left(-\delta,\delta\right)$$. In other words, $$\|h-g\|\geq J/2-\varepsilon$$ for any $$\varepsilon>0$$ and thus, $$\|h-g\|\geq \frac{J}{2}$$.

Claim 2. Let $$g\in \mathcal{D}\left[0,1\right]\setminus \mathcal{C}\left[0,1\right]$$ be a discontinuous càdlàg function -- i.e., it has a positive jump $$J=\left|g(t)-\lim\limits_{x\rightarrow t-} g(x)\right|>0$$ for some $$t$$. Then, $$\mathcal{C}\left[0,1\right]\cap B\left(g,r\right)=\emptyset$$ for any $$r, where $$J$$ is the size of one of the jumps of $$g$$.

Proof. Let $$f\in \mathcal{C}\left[0,1\right]$$. Observe that: $$d(f,g)=\inf_{\lambda\in \Lambda} \left\{ \|\lambda-I\|\vee \|f\circ\lambda - g\| \right\}\geq \inf_{\lambda\in\Lambda}\|f\circ\lambda - g\|.$$

Since $$f\circ \lambda$$ is continuous for any $$\lambda\in\Lambda$$ then, from Claim 1., we have that
$$\|f\circ\lambda - g\|\geq \frac{J}{2}\,\,\,\forall{\lambda\in\Lambda}.$$ Therefore, $$d(f,g)\geq \frac{J}{2}$$, for any $$f\in \mathcal{C}\left[0,1\right]$$. In other words, $$\mathcal{C}\left[0,1\right]\cap B(g,r)=\emptyset$$ for any $$r.

Claim 3. $$\mathcal{D}\left[0,1\right]\setminus\mathcal{C}\left[0,1\right]$$ is dense in $$\mathcal{D}\left[0,1\right]$$.

Proof. Given any $$f\in\mathcal{C}\left[0,1\right]$$, just consider the sequence $$g_n(x):= f(x)+(\frac{1}{n})1_{\left\{x\geq t\right\}}$$ and clearly $$g_n\in\mathcal{D}\left[0,1\right]\setminus \mathcal{C}\left[0,1\right]$$ for all $$n$$. We have $$\|g_n-f\|\overset{n\rightarrow \infty}\longrightarrow 0$$ (which also implies convergence w.r.t. the Skorohod metric).

Answer to the question. Since $$\mathcal{D}\left[0,1\right]\setminus\mathcal{C}\left[0,1\right]$$ is dense in $$\mathcal{D}\left[0,1\right]$$ (Claim 3.), the result follows: i) any (nonempty) open set $$\mathcal{O}$$ in $$\mathcal{D}\left[0,1\right]$$ contains a discontinuous $$g$$; ii) $$B(g,J/3)$$ does not contain any continuous function (Claim 2.), where $$J>0$$ is a jump of $$g$$. Therefore, $$\mathcal{C}\left[0,1\right]$$ is not dense in $$\mathcal{O}$$. This concludes the proof.

• In the last line that is $J/2$ right? Commented Apr 28, 2022 at 14:31
• @Cantor_Set: it should be anything (strictly) smaller than $J/2$ (I have chosen $J/3$). In other words, any continuous $f$ lies outside the ball (induced by the Skorokhod metric) of radius $K$ around $g$ whenever $K<J/2$. Commented Apr 28, 2022 at 15:15
• BTW typesetting suggestion: You can use \| to get $\|$ (rather than ||, which gives $||$). (I think I heard there's yet another code that gives the same result as \| but is semantically preferred for enclosing operators—supposedly, || is to be used for parallel lines and the like—but I don't remember offhand what it is.) Commented Apr 28, 2022 at 16:28
• Ahh yes, it was \lVert and \rVert, so \lVert x \rVert gives $\lVert x \rVert$. Commented Apr 28, 2022 at 16:30
• @WhoKnowsWho: $S$ is nowhere dense in $M$ whenever for any nonempty open subset $O$ of $M$, $S$ is not dense in $O$. The result above follows since any nonempty open set in $D\left[0,1\right]$ contains a discontinuous càdlàg function $g$. I will update as soon as possible to reflect that. In other words, $D\left[0,1\right]\setminus \mathcal{C}\left[0,1\right]$ is dense in $\mathcal{D}\left[0,1\right]$: thus, i) consider any open ball $\mathcal{O}$; ii) it contains a discontinuous càdlàg $g$; iii) $B(g,J/3)$ does not contain any continuous function, where $J$ is the jump of $g$. Commented Apr 28, 2022 at 18:07

Your idea is just opposite. To show that $$C[0, 1]$$ is nowhere dense in $$D[0, 1]$$, you need to show that $$cl(C[0, 1])$$ has empty interior. This means, you would want to start with an arbitrary ball centered at some $$f\in cl(C[0, 1])$$ and then show that that ball is not contained in $$cl(C[0, 1]).$$ This is the idea. How do we implement it?

I think the best way is to follow what Olvier Diaz mentioned in the comment. Start with the observation that the $$cl(C[0, 1])=C[0, 1]$$. This follows from the observation that Skorokhod metric restricted to $$C[0, 1]$$ is the same as $$\|\cdot\|_{\infty}$$.

With this observation, our problem reduces to showing that $$C[0, 1]$$ does not contain a ball (i.e., it has an empty interior). It is a general fact that in a topological vector space, a closed proper subspace is nowhere dense. If you haven't seen this already, there is an easy proof (I am writing it adapted to our case).

Let $$f\in C[0, 1]$$ and let $$\epsilon>0$$ be arbitrary. Consider the ball $$B(\epsilon, f):=\{g\in D[0, 1]: \|f-g\|<\epsilon\}.$$ It suffices to show that there exists $$h\in D[0, 1]\setminus C[0, 1]$$ such that $$h\in B(\epsilon, f)$$. This would show that this ball is not contained in $$C[0, 1]$$. As the epsilon was arbitrary, we conclude that $$C[0, 1]$$ does not contain any ball (and hence any open set).

To construct such a function $$h$$, define $$h(x)=f(x)-sgn(x-1/2)\epsilon/2$$ where $$sgn(x)=1$$ if $$x>0$$ and $$sgn(x)=-1$$ if $$x<0$$ and $$sgn(0)=0$$. Clearly $$h\in D[0, 1]\setminus C[0, 1]$$ and $$\|h-f\|= \epsilon/2<\epsilon.$$ This completes the proof.