Let X = { ( x , y ) ∈ R 2 ∣ y = 0 or x ≥ 0 } and f : X → R defined by f ( x , y ) = x . I want to find C ⊂ X closed such that f ( C ) ⊂ R isn't closed. How to prove that f is a closed map?

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Let   X  =  {  (  x  ,  y  )  ∈            R        2    ∣  y  =  0   or   x  ≥  0  } and   f  :  X  →      R   defined by   f  (  x  ,  y  )  =  x . I want to find   C  ⊂  X closed such that   f  (  C  )  ⊂      R   isn&#039;t closed. How to prove that f is a closed map?

Kaylie Mcdonald · Accepted Answer

Step 1f i understand correctly,   X  =  {  (  x  ,  0  )  :  x  &amp;lt;  0  }  ∪  {  (  x  ,  y  )  :  x  ≥  0  } . So, if we take   C  =  {  (  x  ,  0  )  :  x  &amp;lt;  0  }  ∪  {  (  0  ,  y  )  :  y  ∈      R    } then C is closed because   X  −  C is open in             R        2   but when we apply f (which is the projection map actually, i.e. continuous 1-1 onto), we have   f  (  C  )  =  (  −      ∞    ,  0  ) which is obviously open in       R   .

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