Let $X$ be a complete metric space, $B$ be a (closed) nowhere dense set in $X$, $f:B\rightarrow X$ be a continuous mapping. Is it possible that $f(B)$ is dense in $X$?

I think that is possible. But I can't give a counter-example. Any help will be appreciated.

  • $\begingroup$ It is possible. Can you think of an example if $X=\mathbb R?$ $\endgroup$
    – D. Brogan
    Jul 1 at 15:05
  • $\begingroup$ @D.Brogan Are you sure? I have a proof to the contrary $\endgroup$
    – FShrike
    Jul 1 at 15:33
  • $\begingroup$ (so if you do have a counterexample, I'd like to know because it would show my proof to be false) $\endgroup$
    – FShrike
    Jul 1 at 15:43
  • $\begingroup$ I believe my example has a counterexample @FShrike $\endgroup$ Jul 1 at 17:32
  • $\begingroup$ This is a cool question. Could you provide some context @Ken.Wong? I'm curious where it came from. $\endgroup$ Jul 1 at 17:34

3 Answers 3


$\textbf{Theorem}$: Every compact metric space is a continuous image of the Cantor set $\mathcal{C}$. (See here)

Let us consider the set $[0, 1]$ as euclidean subspace of $(\Bbb{R}, d_{\text{std}}) $.

Then there there exists a continuous onto function $f:\mathcal{C} \to [0, 1]$.

Now $\mathcal{C}\subset [0,1]$ is closed nowhere dense but $f(\mathcal{C})=[0, 1]$ which is a trivial dense subset of $[0,1]$.

Note: $f$ is here the restriction of the cantor function on the cantor set $\mathcal{C}$.

  • $\begingroup$ I am just wondering, what if the nowhere dense set is an infinite-dimensional subspace in Hilbert space, does the conclusion hold? Since the set is not compact anymore. $\endgroup$
    – Ken.Wong
    Jul 2 at 1:58
  • $\begingroup$ @Ken.Wong Of course, as long as your Hilbert space is separable. $\endgroup$ Jul 2 at 4:33
  • 1
    $\begingroup$ @Ken.Wong Consider the Hilbert space $X = \ell_2(\mathbb{N})$, with the nowhere-dense subspace $B$ defined by $x_1 = 0$. The image of $B$ under the left-shift operator $(x_n) \mapsto (x_{n+1})$ is all of $X$, obviously dense. The left-shift is continuous and linear on all of $X$. $\endgroup$ Jul 2 at 16:56

Since $B$ is nowhere dense, it inherits the discrete topology from $X$, which means that any function from $B$ will be continuous. Thus any surjection from $\mathbb{Z}$ onto $\mathbb{Q}$ serves as an example.

But we can do better than this. It is actually possible for $f$ to be continuous on all of $X$, with $f(B)$ still dense! Let $X$ be the positive reals, with $f(x)=x\sin(x)$. Define $$B = \bigcup_{n \in \mathbb{N}} \{2\pi n^3 + \sin^{-1}(\frac{k}{n^4}): k \in \mathbb{N}, k < n^2\}. $$ One can see that $B$ is nowhere dense: every interval of length $2\pi$ contains only finitely many points of $B$. To show $f(B)$ is dense, we need to construct a $b \in B$ with $|b\sin(b) - x| < \epsilon$ for any arbitrary $x, \epsilon > 0$. For large $n$, we have that $\frac{1}{n^2} < \frac{\epsilon}{2}$. For larger $n$, there is some integer $k$ for which $|\frac{2\pi k}{n} - x| < \frac{\epsilon}{2}$. For still larger $n$, we also can guarantee that $k < n^2$. Therefore $B$ contains the point $b = 2\pi n^3 + \sin^{-1}(\frac{k}{n^4})$. Now, $f(b) = b \sin(b) = b \cdot \frac{k}{n^4} = \frac{2\pi k}{n} + \frac{k}{n^4}\sin^{-1}(\frac{k}{n^4})$. Thus,

$$ |b - x| = \bigg| \frac{2\pi k}{n} + \frac{k}{n^4}\sin^{-1}(\frac{k}{n^4}) - x \bigg| \leq \bigg| \frac{2\pi k}{n} - x\bigg| + \bigg|\frac{k}{n^4}\sin^{-1}(\frac{k}{n^4}) \bigg|. $$

The first term is less than $\frac{\epsilon}{2}$ by choice of $k$. The second term is also less than $\frac{\epsilon}{2}$, because $\sin^{-1}(\frac{k}{n^4})$ is bounded by 1, and $\frac{k}{n^4} < \frac{n^2}{n^4} = \frac{1}{n^2} < \frac{\epsilon}{2}$. Therefore $|b - x| < \epsilon$.

There are probably better examples than this. This related post offers a hypothesis which, if true, would give a far more elegant example.

  • $\begingroup$ Thank you for pointing out my error. I was conceptualising $B$ still with the topology on $X$, not as a subspace. My very first thought was using the isomorphisms of $\Bbb Z\cong\Bbb Q$ but I discounted it because I viewed the integers as "not open" - but of course they are, in the subspace $\endgroup$
    – FShrike
    Jul 1 at 17:44

Any space-filling curve gives you an example of a continuous surjective map $$ B=[0,1]\subset X=[0,1]^2\to X. $$ Actually, this situation is quite generic. As long as your metric space $X$ is, say, compact, and contains a nowhere dense subset $B$ homeomorphic to the Cantor set, you get continuous surjective map $B\to X$. (This is a theorem due to Hausdorff.)


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