Given metric spaces ( X , d ) and ( Y , d ′ ) and continuous mapping S and T from X into Y, prove that the set { x ∈ X : S x = T x } is closed in ( X , d ).I've run out of any ideas where I should start

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Given metric spaces   (  X  ,  d  ) and   (  Y  ,      d    ′    ) and continuous mapping   S and   T from   X into   Y, prove that the set   {  x  ∈  X  :  S  x  =  T  x  } is closed in   (  X  ,  d  ).I&#039;ve run out of any ideas where I should start

Dobermann82 · Accepted Answer

Let   E  =  {  x  ∈  X  :  S  x  =  T  x  }. Suppose       x    n    ∈  E and       x    n    →  x.By continuity,   S      x    n    →  S  x and   T      x    n    →  T  x. But   S      x    n    =  T      x    n   for all   n  ≥  1, hence,   S  x  =  T  x and this means   x  ∈  E. This implies   E  =  {  x  ∈  X  :  S  x  =  T  x  } is closed.

Given metric spaces ( X , d ) and ( Y , d ′ </ms

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