Two spaces are not homeomorphic

[beetle2]

Prove that no continuous surjective function $f : ]0; 1] \rightarrow R$can be injective.


My questions is can I use a proof by contradiction and assume that there is a injection

Then I can use the fact that if there was an injection ie there's a bijective function then the two spaces would be homeomorphic to each other.

And then show the the two spaces are not homeomorphic so impossible?

 

[Dickfore]

If there's a bijection, they would be isomorphic as far as I can tell. I think that it has something to do with the continuity and that the segment is closed.

 

[quasar987]

Prove that no continuous surjective function $f : ]0; 1] \rightarrow R$can be injective.


My questions is can I use a proof by contradiction and assume that there is a injection

Then I can use the fact that if there was an injection ie there's a bijective function then the two spaces would be homeomorphic to each other.

And then show the the two spaces are not homeomorphic so impossible?

Of course.

 

[lavinia]

Prove that no continuous surjective function $f : ]0; 1] \rightarrow R$can be injective.


My questions is can I use a proof by contradiction and assume that there is a injection

Then I can use the fact that if there was an injection ie there's a bijective function then the two spaces would be homeomorphic to each other.

And then show the the two spaces are not homeomorphic so impossible?

the inverse image of the positive integers has a limit point

 

[beetle2]

Then I will try and show that both spaces are homeomorhic.

 

[Dickfore]

Suppose that f is continuous and injective. Maybe it's easiest to prove that it cannot be a surjection, i.e. there is at least one real number which is not the image of any point on the segment [0, 1].

 

[beetle2]

But the segment is (0.1]

 

[Dickfore]

You only need one boundary point contained in the finite region, so i think you can still do it.

 

[beetle2]

So If I assume that f is continuous and injective:

take
$a \in X a = 1$such that $f(a) \in Y$
And as $f$is continuous and injective I can find for any $\epsilon > 0$
a real element $y \in Y$ such that the inverse $f^{-1}(y+ \epsilon) - f^{-1}(y) + f^{-1}(y) - f^{-1}(y-\epsilon)$ is in $X$
How ever since $f^{-1}f(a)$ is a boundary point this is impossible

showing there is at leat one real number which is not the image of any point on the segment (0, 1].

 

[quasar987]

Prove that no continuous surjective function $f : ]0; 1] \rightarrow R$can be injective.


My questions is can I use a proof by contradiction and assume that there is a injection

Then I can use the fact that if there was an injection ie there's a bijective function then the two spaces would be homeomorphic to each other.

And then show the the two spaces are not homeomorphic so impossible?

I'm sorry, I read your post a little too fast. Your second sentence contains a false statement: the existence of a continuous bijective function does not imply the existence of a homeomorphism.

To solve this problem, try supposing that there is a continuous injective and surjective function f:]0,1]-->R. Then f([½,1])=[a,b] for some real numbers a and b. Suppose that f(1)=b (the case f(1)=a is similar). Show that no x>b is hit by f. In other words, $f^{-1}(]b,\infty[)=\emptyset$ and thus f is not surjective: a contradiction.

 

[boneill3]

I appreciate all your help guys.

 

[Jamma]

Your map can't exist in the first place; the image of a compact subset under a continuous map is compact, and the reals are not compact.

Other than that, I suppose you could also say that if your map was injective, then since bijective maps from compact spaces to Hausdorff spaces are homeomorphisms, then you would have a contradiction since [0,1] is not homeomorphic to the reals, since for example [0,1] is compact; although this is the point I made in the first paragraph.

 

[Dickfore]

Your map can't exist in the first place; the image of a compact subset under a continuous map is compact, and the reals are not compact.

Prove it.

 

[Office_Shredder]

The domain isn't compact anyway (it doesn't contain 0). As for proving the reals are NOT compact, that's easy since all you need to do is come up with a counterexample to the definition.

quasar has the right idea... basically, for there to be a continuous bijection it has to be monotone (increasing or decreasing). Then f(1) taking a finite value causes serious problems

 

[Landau]

A subset of R^n is compact iff it is closed and bounded. ]0,1] is not compact because it is not closed. R is not compact because it is not bounded.

 

[lavinia]

the negative integers -m map to an infinite decreasing or increasing sequence in (0,1] that either has a limit point or converges to 0. If there is a limit point the inverse can not be a homeomorphism. but then the positive integers must have a limit point.

 

[Bacle]

Maybe you can use the fact that a continuous bijection between [0,1] and R
would be (is) a homeo. (cont. bijection, compact, Hausd. , yada yada)

Then, if h is that homeo., the restriction:

h^{^}:(0,1]-->R-{f(0)} is also a homeo.

but R-{f(0)} is disconnected, and (0,1] (as a subspace) is not.

 

[quasar987]

This looks like an argument against the existence of a homeomorphism between R and [0,1] (which is direct by compactness by the way). The OP was trying to find an argument against the existence of a homeo between R and (0,1].

 

[Bacle]

I was not claiming this is a solution; just mentioning
a result that seemed closed to what the OP was asking
for; I thought it may help give him an idea.

 

[Jamma]

Sorry, I thought that the ]0,1] was a typo to mean [0,1], in which case I'd obviously have been correct. I've never seen this (very weird) notation, and would use (0,1].

Looking at it though, it does make a bit of sense, its like the brackets are indicating visually where the boundaries lie on the real line, and it can be annoying having things like (0,1) around when there are also 2-tuples.

 

[quasar987]

The notation is universal in french books. I've seen it on rare occasions in english books, perhaps by authors trying to popularize it.

 

[some_dude]

Since f is continuous, the preimage $$A = f^{-1}((f(1), +\infty))$$ must be open in (0, 1]. Ditto for $$B = f^{-1}((-\infty, f(1))$$.

Then $$(0, 1) \subset A \cup B$$ and the surjectivity assures neither A nor B are empty. And this along with their openness implies $$A \cap B \ne \emptyset$$, which implies $$A \cup B$$ is connected, and so $$f(A \cup B)$$ is connected (and equal the whole real line), and therefore $$f(1) \in f(A \cup B)$$. So f is not injective.