Question clarification.

Question

This is an exercise question:

"Let $V$ be an open subset of $\mathbb{R}^3$. Does there exist a continuous injective function $f:V \rightarrow \mathbb{R}$?"

What exactly does it mean?

Is the question: "Given an open subset of $\mathbb{R}^3$, does there always exist a continuous injective function $f:V \rightarrow \mathbb{R}$?"? $$ \forall V\subseteq \mathbb{R}^3:\ V \text{ is open } \Rightarrow \exists f:\ V \rightarrow \mathbb{R} \text{ continuous and injective} $$

Or is it maybe: "Does there exist a continuous injective function $f:V \rightarrow \mathbb{R}$? where $V$ is an open subset of $\mathbb{R}^3$.

$$ \exists f:\ V \rightarrow \mathbb{R} \text{ continuous and injective}, V\subseteq \mathbb{R}^3:\ V \text{ is open } $$

I honestly do not have a clue. How I would go about (dis)proving this would very much depend on my understanding.

Probably neither. I'd modify the first one to say, "Given an open subset of R^3, when does there exist a ...." The answer may be always, or never, or something in between. — muaddib, Jun 11 '15 at 10:35
You say "$V\subseteq \mathbb{R}$" twice, whereas $V$ is a subset of $\mathbb R^3$. (And therefore there can never exist such a function, if $V$ is non-empty.) — TonyK, Jun 11 '15 at 10:39
You're right, my mistake. I've corrected this to $\mathbb{R}^3$. — Syd Kerckhove, Jun 11 '15 at 10:40

AlexR · Accepted Answer · 2015-06-11T11:51:42.443

1

The statement can be reformulated to

Let $V\subset \mathbb R^3$ be open. Under what circumstances does there exist a continuous injective function $f:V\to\mathbb R$?

If you really want to make it formal, you can write

Find a property $P$ such that $$\forall V \subset \mathbb R^3 \text{ open}: P(V) \implies \exists f:V\to \mathbb R\text{ continuous and injective}$$

The answer would then be $P(V) := V = \emptyset$.

If $V$ is the empty set, there is such an $f$ (called the empty function). It doesn't map anything and since $\mathrm{dom}(f) = \emptyset$, all statements about preimages of $f$ are vacuously true, hence $f$ is injective.

If $V$ isn't empty, it contains some open ball $B_\epsilon(x_0) \subset \mathbb R^3$. Since there is no injective map from $B_1(0)$ to $\mathbb R$ (and any $B_\epsilon(x_0)$), there cannot be an injective map from $V$ to $\mathbb R$.

Thus, the answer to the question is:

If $V\subset \mathbb R^3$ is open, there exists a continuous function $f:V\to\mathbb R$ if and only if $V=\emptyset$.

edited Jun 11 '15 at 11:51

answered Jun 11 '15 at 10:46

AlexR

24,905

1

Could you show me why there is no injective map from $B(0,1) \subseteq \mathbb{R}^3$ to $\mathbb{R}$? Maybe a reference to a proof? In $\mathbb{R}$ it seems to be OK: http://math.stackexchange.com/questions/200180/is-there-a-bijective-map-from-0-1-to-mathbbr – Syd Kerckhove Jun 11 '15 at 10:50
@SydKerckhove This is Borsuk-Ulam Theorem. – Crostul Jun 11 '15 at 11:18
Well that seems to be far beyond the reach of an 'intro to analysis' student... – Syd Kerckhove Jun 11 '15 at 11:20
@SydKerckhove Unfortunately, your statement is basically equivalent to the Borsuk-Ulam theorem. Maybe you're only expected to state this without proof? – AlexR Jun 11 '15 at 11:42
Definitely not. We're not even expected to have ever heard of that theorem. I have found a proof at the level of an intro to analysis student: http://math.stackexchange.com/questions/1321194/continuous-injective-function-from-open-part-of-mathbbr3-to-mathbbr – Syd Kerckhove Jun 11 '15 at 11:43
@SydKerckhove Nice. This is a proof of said theorem. By the way, you should have answered this here and not posted a duplicate question. – AlexR Jun 11 '15 at 11:46
Acually, this question was about the formulation, not about the proof... – Syd Kerckhove Jun 11 '15 at 11:47
@SydKerckhove Sorry for that. I'll add an answer to this as well. – AlexR Jun 11 '15 at 11:48
Your new edited answer is perfect. Thank you very much. – Syd Kerckhove Jun 11 '15 at 12:17

score -1 · Answer 2 · answered Jun 11 '15 at 11:36

Use the following fact:

There is no locally injective continuous function $f:[0,1] \to \Bbb{R}$ such that $f(0)=f(1)$.

Now, since $V$ is open, you can build a closed curve (for example, the equator of a sphere) inside some ball contained in $V$. This means that you are able to find some $\gamma: [0,1] \to V$ with $\gamma(0)=\gamma(1)$, which parametrizes your curve. You can suppose WLOG that $\gamma|_{[0,1)}$ is injective.

If $f: V \to \Bbb{R}$ is injective and continuous, then $f \circ \gamma$ is a locally injective continuous function and $f \circ \gamma(0) = f \circ \gamma(1)$: a contradiction.

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