Topology questions

**ogele** · March 1st 2010, 06:14 PM

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

**mabruka** · March 1st 2010, 06:48 PM

a) $\sin (\pi x)<br />$
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

**Drexel28** · March 1st 2010, 07:00 PM

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.

**mabruka** · March 2nd 2010, 04:46 AM

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

**ogele** · March 2nd 2010, 05:53 AM

Thanks for answering the question Mabruka and Drexel28

**tonio** · March 2nd 2010, 10:48 AM

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

**Drexel28** · March 2nd 2010, 03:27 PM

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.

**Plato** · March 2nd 2010, 03:47 PM

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?

**Drexel28** · March 2nd 2010, 04:36 PM

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.

**Plato** · March 2nd 2010, 04:49 PM

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?

**Drexel28** · March 2nd 2010, 04:54 PM

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?

**tonio** · March 2nd 2010, 06:47 PM

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

**Drexel28** · March 2nd 2010, 07:08 PM

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.

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