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determine whether or not a subset is closed or open:

(a) For $X=\Bbb R^2$ and $d$ the Euclidean metric on $\Bbb R^2$:

  1. $A_1=${$(x,y): x^2+y^2 <1$} $\cup $ {$(1,0)$}.

  2. $A_2=${$(x,0): 0 < x < 1$}.

(b) For $X=${all continuous functions $f: [0,1]\to [0,1]$ } with the metric $d(f,g) = \sup_{x\in [0,1]}|f(x) - g(x)|$:

  • $A_3=${$f\in X : f(0)=f(1)$}.
Community
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Jhwana
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1 Answers1

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For $A_1$; look at open balls with center at $(1,0)$. What fails? Then find a sequence of poits in $A_1$ such that $\lim a_n \notin A_1$. That shows $A_1$ is nor open, neither closed.

For $A_2$, note we can find again a sequence of points for which $\lim a_n\notin A_2$. This shows $A_2$ is not closed, not that it is open. Now look at the open balls with points centers of $A_2$. What fails?

For the third, consider the function $$G:X\to \Bbb R$$ defined by $G(f)=f(1)-f(0)$. Then $A_3=f^{-1}( \{ 0\})$. Is it continuous? If you can prove it is continuous, then $\{0\}$ being closed in $\Bbb R$ implies $A_3$ is what in $X$?

Pedro
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  • for $A_2$: $a_n=(\frac 1n,0)$ and $lim$ $ a_n=0\notin A_2$, so $A_2$ is OPEN.(Is this correct?) – Jhwana Mar 21 '13 at 18:13
  • @Fayz I don't like how I hinted you. Let me rephrase. – Pedro Mar 21 '13 at 18:19
  • $lim$ $a_n =(0,0)∉A_2$ – Jhwana Mar 21 '13 at 18:27
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    @Fayz Right. So $A_2$ is not closed. Remember that open and closed are not mutually exclusive terms in topology, as they are in everyday speech. It would be a good exercise that you construct some sets that are neither open nor closed; open, but not closed; closed, but not open; and both open and closed. – Pedro Mar 21 '13 at 18:29