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 Post subject: set inclusion
PostPosted: Thu, 13 Oct 2011 16:55:16 UTC 
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hi, can you give me an idea of this?

let (\Omega, d) and (\Omega \prime , d\prime ) be two metric spaces, let f:\Omega \mapsto \Omega \prime continuous, for A\subseteq \Omega

f(\overline A)\subseteq \overline{f(A)}

give an example where the inclusion is strict.

I´m not looking for someone to give me the answer, just an idea, thanks


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 Post subject: Re: set inclusion
PostPosted: Thu, 13 Oct 2011 17:10:02 UTC 
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ender wrote:
hi, can you give me an idea of this?

let (\Omega, d) and (\Omega \prime , d\prime ) be two metric spaces, let f:\Omega \mapsto \Omega \prime continuous, for A\subseteq \Omega

f(\overline A)\subseteq \overline{f(A)}

give an example where the inclusion is strict.

I´m not looking for someone to give me the answer, just an idea, thanks


You can find them inside \mathbb{R}.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: set inclusion
PostPosted: Thu, 13 Oct 2011 17:58:56 UTC 
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Let f:R->R be a function where R is the reals.

Okay, the closure of f(A) is basically points approached by y-values of the graph.
Let y \in \overline{f(A)}. Suppose y \in f(\overline{A}), then f(x)=y for some x in \overline{A}.

So if you don't want y in f(\overline{A}), you can't have f(x)=y for any x in A and you can't have f(x)=y for any x real with x_n \to x for some sequence x_n \in A.

Thus, a continuous function with a horizontal asymptote it never crosses should be fine, there are tons of functions like this.


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