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The statement is: The unit interval can never be made into a topological group under any multiplication.

$\textbf{HINT:}$ For G to be a topological group,then for every two elements $x,y \in G$ ,there exists a homeomorphism $h : G \rightarrow G$ such that $h(x) = y$.As I know the homeomorphism will be the right translation by $x^{-1}y$.

But I am stuck in how to use this hint to prove the above statement.

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The statement

For all $x,y \in X$ there is a homeomorphism $h : X \to X$ such that $h(x) = y$

is the definition of "$X$ is a homogeneous space". It is the fomalisation of the idea that all points of $X$ "look/behave the same", topologically. Indeed translations show that a topological group is always homogeneous.

But for $X = [0,1]$ we can prove that $X$ is not homogeneous.

Take $x = 0$ and $y = \frac{1}{2}$.

If $h: [0,1] \to [0,1]$ would be a homeomorphism with $h(0) = \frac{1}{2}$, then the restriction $h: [0,1]\setminus \{0\} \to [0,1]\setminus \{\frac{1}{2}\}$ is also a homeomorphism. But $(0,1]$ is connected and $[0,\frac{1}{2}) \cup (\frac{1}{2}, 1]$ is not, so we have a contradiction, and so $h$ cannot exist.

As $[0,1]$ is not homogeneous, it cannot be made into a topological group.

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Hint: in a topological group, if removal of a point $x$ disconnects the underlying topological space, then so does removal of $xy$ for any $y$.

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  • $\begingroup$ I have just started studying Topological groups and I didn't get you. $\endgroup$ – Sumit Mittal Feb 24 '18 at 12:40
  • $\begingroup$ $[0, 1] \setminus \{1\} = [0, 1)$ is connected while $[0, 1] \setminus \{1/2\} = [0, 1/2) \cup (1/2, 1]$ is not. Hence there is no homeomorphism of $[0, 1]$ to $[0, 1]$ mapping $1$ to $1/2$. Now use the HINT in your question. $\endgroup$ – Rob Arthan Feb 24 '18 at 15:23

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