Let X be a locally compact (not compact) Hausdorff space. Let Y = X ∪ { 0 } be the one point compactification of X. Now define a set C 0 ( X ) by: C 0 ( X ) := { f ∈ C ( X ) : f vanishes at infinity } .Now we say that f ∈ C ( X ) vanishes at infinity if for each ε &gt; 0 the set { ω ∈ X : | f ( ω ) | ≥ ε } is compact. Now consider C ( Y ), the set of all continuous functions from Y to C . Now there is a statement in the middle of a proof that, a function in C ( Y ) belongs to C 0 ( X ) if and only if f ( 0 ) = 0. I am not able to prove this statement.My approach: ′′ ″ ⇐ ″ ′′ Let f ∈ C ( Y ) and f ( 0 ) = 0. Nowf is continuous at 0. Then for each ε &gt; 0 there exists a neighborhood U 0 of 0 such that | f ( x ) | &lt; ε for all x ∈ U 0 . Now Y is compact and U 0 is open hence Y ∖ U 0 is closed hence compact. Now I am not able to proceed from here and also the ′′ ″ ⇒ ″ ′′ direction.

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Let X be a locally compact (not compact) Hausdorff space. Let   Y  =  X  ∪  {  0  } be the one point compactification of X. Now define a set       C    0    (  X  ) by:      C    0    (  X  )  :=      {    f    ∈    C    (    X    )    :    f     vanishes at infinity    }    .Now we say that   f  ∈  C  (  X  ) vanishes at infinity if for each   ε  &amp;gt;  0 the set   {  ω  ∈  X  :      |    f  (  ω  )      |    ≥  ε  } is compact. Now consider   C  (  Y  ), the set of all continuous functions from Y to       C  . Now there is a statement in the middle of a proof that, a function in   C  (  Y  ) belongs to       C    0    (  X  ) if and only if   f  (  0  )  =  0. I am not able to prove this statement.My approach:  ′′           ″        ⇐    ″     ′′  Let   f  ∈  C  (  Y  ) and   f  (  0  )  =  0. Nowf is continuous at 0. Then for each   ε  &amp;gt;  0 there exists a neighborhood       U    0   of 0 such that       |    f  (  x  )      |    &amp;lt;  ε for all   x  ∈      U    0  . Now Y is compact and       U    0   is open hence   Y  ∖      U    0   is closed hence compact. Now I am not able to proceed from here and also the ′′          ″        ⇒    ″  ′′ direction.

Mariam Hickman · Accepted Answer

(  ⇐) To continue, observe that   {  ω  ∈  X  :      |    f  (  ω  )      |    ≥  ε  }  ⊂  X  ∖      U    0  .(  ⇒) Assume now that   f  ∈  C  (  Y  ) and that   f (or more appropriately, its restriction to   X) vanishes at infinity. We want to show that   f  (  0  )  =  0. Using continuity, it suffices to show that for each   ε  &amp;gt;  0, there is some open neighborhood   U  ⊂  Y of 0 such that       |    f  (  x  )      |    &amp;lt;  ε for all   x  ∈  U,   x  ≠  0.

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