Linked Questions

2
votes
4answers
432 views

Non-existence of bijective, continuous function from $(0,1)$ to $[0,1]$ [duplicate]

The problem : Give example of a continuous, onto function from $(0,1)$ to $[0,1]$. Is it possible for such a function to be one-one? My partial solution : For the $1^{st}$ part of the question, I ...
-1
votes
2answers
429 views

Is there a bijective, continuous mapping from $\mathbb{R}$ to the closed interval $[0,1]$? [duplicate]

i can't find a bijective, continuous map from $\mathbb{R}$ to the closed interval $[0,1]$. Give an example. If not bijective then what is the difference between cardinal no of $(0,1)$ and $[0,1]$ ?
3
votes
3answers
127 views

No bijective continuous $f:(0,1) \to [0,1]$ using facts from topology? [duplicate]

In the context of learning topology, I'm triyng to see how to show that there is no bijective continuous function $f:(0,1) \to [0,1]$ using some basic topology facts such as the definition of ...
0
votes
1answer
177 views

Whether there is a continuous bijection from $(0,1)$ to closed interval $[0,1]$. [duplicate]

Is there a continuous bijection from open interval $(0,1)$ to $[0,1]$. The answer is not. How to prove? I think it may proceed by contradiction and apply open mapping theorem. However, $(0,1)$ is ...
3
votes
2answers
134 views

Does there exists a onto continuos function $f\colon (0,1) \to[0,1]$ which is one to one . [duplicate]

Iam trying to solve this problem , i can visualize graphically it is not possible. iam trying to do the proof by contradiction. if f:(0,1) to [0,1] is a continuos onto function then i have to prove ...
0
votes
1answer
54 views

Does there exist any continuous bijection between [0,1] and (0,1) and between [0,1] and IR? [duplicate]

We know that there are bijections between $[0,1]$, $(0,1)$ and $\mathbb{R}$. But my question is can we obtain a continuous bijection between $[0,1]$ and $(0,1)$, and between $[0,1]$ and $\mathbb{R}$? ...
148
votes
8answers
36k views

How to define a bijection between $(0,1)$ and $(0,1]$?

How to define a bijection between $(0,1)$ and $(0,1]$? Or any other open and closed intervals? If the intervals are both open like $(-1,2)\text{ and }(-5,4)$ I do a cheap trick (don't know if that'...
9
votes
6answers
8k views

Mapping the Real Line to the Unit Interval

What is a continuous mapping of the real line $(-\infty, \infty)$ to the interval $[0, 1]$? I'm trying out logs and exponentials but they don't seem to work?
15
votes
5answers
12k views

Continuous function from $(0,1)$ onto $[0,1]$

While revising, I came across this question(s): A) Is there a continuous function from $(0,1)$ onto $[0,1]$? B) Is there a continuous one-to-one function from $(0,1)$ onto $[0,1]$? (clarification: ...
10
votes
2answers
3k views

Is there a continuous function from $[0, 1]$ onto $(0, 1)$?

If there is none, why? And for the other side, what about open set $(0, 1)$ to closed set $[0, 1]$ with a continuous function? Thanks
5
votes
4answers
5k views

Need a hint: prove that $[0, 1]$ and $(0, 1)$ are not homeomorphic

I need a hint: prove that $[0, 1]$ and $(0, 1)$ are not homeomorphic without referring to compactness. This is an exercise in a topology textbook, and it comes far earlier than compactness is ...
8
votes
2answers
5k views

Bijection from (0,1) to [0,1)

I'm trying to solve the following question: Let $f:(0,1)\to [0,1)$ and $g:[0,1)\to (0,1)$ be maps defined as $f(x)=x$ and $g(x)=\frac{x+1}{2}$. Use these maps to build a bijection $h:(0,1)\...
3
votes
1answer
3k views

Why are there no continuous one-to-one functions from (0, 1) onto [0, 1]? [closed]

I do not understand the justification of why III is false, could anyone clarify this for me please? Which of the following statements are true about the open interval $(0,1)$ and the closed ...
7
votes
1answer
1k views

Is there exist a homemoorphism between either pair of $(0,1),(0,1],[0,1]$

As the topic is there exist a homeomorphism between either pair of $(0, 1),(0,1],[0,1]$
3
votes
3answers
323 views

Is there a bijection from $(-\infty,\infty)$ to $[0,1]$?

Previous questions established, for example, that a continuous bijection $f:(0,1)\to [0,1]$ does not exist, but the proof relied on continuity. Clearly, with continuity similar proofs can be invoked ...

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