# Math Help - Topology questions

1. ## Topology questions

I need help with these questions

1. Does there exist a continuous function from (0,1) onto [0,1]
2. Does there exist a continuous function from (-1,1) onto R
3. Does there exist a continuous function from R onto (-1, 1)
4. Does there exist a continuous function from R onto Q
5. Does there exist a continuous function from Q onto R

where R = real numbers
and Q = rational numbers

2. a) $\sin (\pi x)
$

b) $\tan (\frac{\pi}{2}x)$

c) $\frac{2}{\pi}\arctan (x)$

d) no, since R is connected and Q is disconnected

because the continuous image of a connected set must be connected

e)hmmm

3. Originally Posted by ogele
5. Does there exist a continuous function from Q onto R

where R = real numbers
and Q = rational numbers
Of course not. A general theorem (requiring Zorn's lemma) implies that if $f:X\mapsto Y$ is surjective then there exists an injection $g:Y\mapsto X$. Does there exist an injection $g:\mathbb{R}\mapsto \mathbb{Q}$.

And since this thread is labeled topology I suppose that I can impose any topology on these sets.

Give $\mathbb{Q}$ the discrete topology and $\mathbb{R}$ the indiscrete. Then any mapping $f:\mathbb{Q}\mapsto\mathbb{R}$ is continuous.

4. I forgot about the label "topology"!!

Ogele, since you did not specified the topologies it is understanded we have the usual topology in both the domain and range

5. ## Topology Questions

Thanks for answering the question Mabruka and Drexel28

6. Originally Posted by Drexel28
Of course not. A general theorem (requiring Zorn's lemma) implies that if $f:X\mapsto Y$ is surjective then there exists an injection $g:Y\mapsto X$. Does there exist an injection $g:\mathbb{R}\mapsto \mathbb{Q}$.

And since this thread is labeled topology I suppose that I can impose any topology on these sets.

Give $\mathbb{Q}$ the discrete topology and $\mathbb{R}$ the indiscrete. Then any mapping $f:\mathbb{Q}\mapsto\mathbb{R}$ is continuous.

Or simpler, if we already got AC into the game: if there exists a surjective $\mathbb{Q}\mapsto \mathbb{R}$ then $\aleph_0=Card(\mathbb{Q})\geq Card(\mathbb{R})=2^{\aleph_0}$ , which of course is absurd.

Tonio

7. Originally Posted by tonio
Or simpler, if we already got AC into the game: if there exists a surjective $\mathbb{Q}\mapsto \mathbb{R}$ then $\aleph_0=Card(\mathbb{Q})\geq Card(\mathbb{R})=2^{\aleph_0}$ , which of course is absurd.

Tonio
Well, even simpler (without AC). We can use the well-ordering principle to prove that if $f:\mathbb{N}\mapsto E$ is surjective then $E$ is countable.

8. Originally Posted by Drexel28
Well, even simpler (without AC). We can use the well-ordering principle to prove that if $f:\mathbb{N}\mapsto E$ is surjective then $E$ is countable.
Do you understand that
Axiom of Choice $\Rightarrow$ Hausdroff’s Maximal Principle $\Rightarrow$ Zorn’a Lemma $\Rightarrow$ well-ordering theorem $\Rightarrow$ Axiom of Choice?

9. Originally Posted by Plato
Do you understand that
Axiom of Choice $\Rightarrow$ Hausdroff’s Maximal Principle $\Rightarrow$ Zorn’a Lemma $\Rightarrow$ well-ordering theorem $\Rightarrow$ Axiom of Choice?
I'm kind of shocked you'd ask that. But, yes I do. I thought that the well-ordering principle of the natural numbers did not require the axiom of choice...but I could be wrong.

10. Originally Posted by Drexel28
A general theorem (requiring Zorn's lemma) implies that if $f:X\mapsto Y$ is surjective then there exists an injection $g:Y\mapsto X$. Does there exist an injection $g:\mathbb{R}\mapsto \mathbb{Q}$.
Did you not post the above?
So why are you surprised?

11. Originally Posted by Plato
Did you not post the above?
So why are you surprised?
Yes, but you can prove that if a set is the surjective image of a countable set it's countable using only the well-ordering principle of the natural numbers. Is that equivalent to the AOC?

12. Originally Posted by Drexel28
Yes, but you can prove that if a set is the surjective image of a countable set it's countable using only the well-ordering principle of the natural numbers. Is that equivalent to the AOC?

I think this may be a confusion. The well ordering principle is usually taken to be: any non-empty subset of the naturals has a first element (in the usual, natural ordering), and this is logically equivalent to the (weak) principle of induction.

Perhaps what Plato assumed, and this is what I thought at first, is Zermelo's well ordering theorem: every non-empty set can be well-ordered , and this in ZF is equivalent to Zorn's lemma and to AC.

Tonio

13. Originally Posted by tonio
I think this may be a confusion. The well ordering principle is usually taken to be: any non-empty subset of the naturals has a first element (in the usual, natural ordering), and this is logically equivalent to the (weak) principle of induction.

Perhaps what Plato assumed, and this is what I thought at first, is Zermelo's well ordering theorem: every non-empty set can be well-ordered , and this in ZF is equivalent to Zorn's lemma and to AC.

Tonio
Maybe. I meant the well-ordering of the naturals. That's all you need.