Prove that the space of continuous functions $C(X,Y)$ with the topology of compact convergence is Hausdorff.











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Let $X$ be a topological space and $Y$ be a metric space. Prove that the space of continuous functions $C(X, Y )$ with the topology of compact convergence is Hausdorff.



The compact convergence topology is generated by the collection of sets $B_C(f,ϵ)={g∈Y^X: text{sup }_{x∈C} d(f(x),g(x))<ϵ}$, with compact $C⊂X$.



I know to show a topology is Hausdorff we need to show that for any two distinct elements in the topology we can find neighborhoods of those elements that do not intersect, but I am not sure how to do this for the compact convergence topology. The functions and supremum are throwing me off.










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    Let $X$ be a topological space and $Y$ be a metric space. Prove that the space of continuous functions $C(X, Y )$ with the topology of compact convergence is Hausdorff.



    The compact convergence topology is generated by the collection of sets $B_C(f,ϵ)={g∈Y^X: text{sup }_{x∈C} d(f(x),g(x))<ϵ}$, with compact $C⊂X$.



    I know to show a topology is Hausdorff we need to show that for any two distinct elements in the topology we can find neighborhoods of those elements that do not intersect, but I am not sure how to do this for the compact convergence topology. The functions and supremum are throwing me off.










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      up vote
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      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $X$ be a topological space and $Y$ be a metric space. Prove that the space of continuous functions $C(X, Y )$ with the topology of compact convergence is Hausdorff.



      The compact convergence topology is generated by the collection of sets $B_C(f,ϵ)={g∈Y^X: text{sup }_{x∈C} d(f(x),g(x))<ϵ}$, with compact $C⊂X$.



      I know to show a topology is Hausdorff we need to show that for any two distinct elements in the topology we can find neighborhoods of those elements that do not intersect, but I am not sure how to do this for the compact convergence topology. The functions and supremum are throwing me off.










      share|cite|improve this question













      Let $X$ be a topological space and $Y$ be a metric space. Prove that the space of continuous functions $C(X, Y )$ with the topology of compact convergence is Hausdorff.



      The compact convergence topology is generated by the collection of sets $B_C(f,ϵ)={g∈Y^X: text{sup }_{x∈C} d(f(x),g(x))<ϵ}$, with compact $C⊂X$.



      I know to show a topology is Hausdorff we need to show that for any two distinct elements in the topology we can find neighborhoods of those elements that do not intersect, but I am not sure how to do this for the compact convergence topology. The functions and supremum are throwing me off.







      general-topology






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      asked Nov 16 at 0:30









      frostyfeet

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          Suppose $f neq g$. We know that there must be some $p in X$ with $f(p) neq g(p)$.
          Define $varepsilon = frac{d(f(p),g(p))}{2} >0$, where $d$ is the metric on $Y$.



          Now define $C={p}$, which is finite, so compact. A supremum of $d(f(x), g(x))$ with $x in C$ is then just the single value $d(f(p), g(p))$.



          Then $f in B_C(f,varepsilon) = {h in Y^X: d(f(p), h(p)) < varepsilon}$ and $g in B_C(g,varepsilon) = {h in Y^X: d(g(p), h(p)) < varepsilon}$.



          If $h in B_C(f, varepsilon) cap B(g, varepsilon)$ existed, then both



          $$d(h(p), f(p)) < varepsilon text{, and } d(h(p), g(p)) < varepsilon$$



          and so $$d(f(p), g(p)) le d(f(p), h(p)) + d(h(p), g(p)) < 2varepsilon = d(f(p), g(p)$$



          which is a contradiction. So $B_C(f, varepsilon) cap B(g, varepsilon) = emptyset$ and $C(X,Y)$ is Hausdorff.






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            Hint: Singleton sets ${x}$ are always compact






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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote



              accepted










              Suppose $f neq g$. We know that there must be some $p in X$ with $f(p) neq g(p)$.
              Define $varepsilon = frac{d(f(p),g(p))}{2} >0$, where $d$ is the metric on $Y$.



              Now define $C={p}$, which is finite, so compact. A supremum of $d(f(x), g(x))$ with $x in C$ is then just the single value $d(f(p), g(p))$.



              Then $f in B_C(f,varepsilon) = {h in Y^X: d(f(p), h(p)) < varepsilon}$ and $g in B_C(g,varepsilon) = {h in Y^X: d(g(p), h(p)) < varepsilon}$.



              If $h in B_C(f, varepsilon) cap B(g, varepsilon)$ existed, then both



              $$d(h(p), f(p)) < varepsilon text{, and } d(h(p), g(p)) < varepsilon$$



              and so $$d(f(p), g(p)) le d(f(p), h(p)) + d(h(p), g(p)) < 2varepsilon = d(f(p), g(p)$$



              which is a contradiction. So $B_C(f, varepsilon) cap B(g, varepsilon) = emptyset$ and $C(X,Y)$ is Hausdorff.






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                up vote
                1
                down vote



                accepted










                Suppose $f neq g$. We know that there must be some $p in X$ with $f(p) neq g(p)$.
                Define $varepsilon = frac{d(f(p),g(p))}{2} >0$, where $d$ is the metric on $Y$.



                Now define $C={p}$, which is finite, so compact. A supremum of $d(f(x), g(x))$ with $x in C$ is then just the single value $d(f(p), g(p))$.



                Then $f in B_C(f,varepsilon) = {h in Y^X: d(f(p), h(p)) < varepsilon}$ and $g in B_C(g,varepsilon) = {h in Y^X: d(g(p), h(p)) < varepsilon}$.



                If $h in B_C(f, varepsilon) cap B(g, varepsilon)$ existed, then both



                $$d(h(p), f(p)) < varepsilon text{, and } d(h(p), g(p)) < varepsilon$$



                and so $$d(f(p), g(p)) le d(f(p), h(p)) + d(h(p), g(p)) < 2varepsilon = d(f(p), g(p)$$



                which is a contradiction. So $B_C(f, varepsilon) cap B(g, varepsilon) = emptyset$ and $C(X,Y)$ is Hausdorff.






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                  up vote
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                  accepted







                  up vote
                  1
                  down vote



                  accepted






                  Suppose $f neq g$. We know that there must be some $p in X$ with $f(p) neq g(p)$.
                  Define $varepsilon = frac{d(f(p),g(p))}{2} >0$, where $d$ is the metric on $Y$.



                  Now define $C={p}$, which is finite, so compact. A supremum of $d(f(x), g(x))$ with $x in C$ is then just the single value $d(f(p), g(p))$.



                  Then $f in B_C(f,varepsilon) = {h in Y^X: d(f(p), h(p)) < varepsilon}$ and $g in B_C(g,varepsilon) = {h in Y^X: d(g(p), h(p)) < varepsilon}$.



                  If $h in B_C(f, varepsilon) cap B(g, varepsilon)$ existed, then both



                  $$d(h(p), f(p)) < varepsilon text{, and } d(h(p), g(p)) < varepsilon$$



                  and so $$d(f(p), g(p)) le d(f(p), h(p)) + d(h(p), g(p)) < 2varepsilon = d(f(p), g(p)$$



                  which is a contradiction. So $B_C(f, varepsilon) cap B(g, varepsilon) = emptyset$ and $C(X,Y)$ is Hausdorff.






                  share|cite|improve this answer












                  Suppose $f neq g$. We know that there must be some $p in X$ with $f(p) neq g(p)$.
                  Define $varepsilon = frac{d(f(p),g(p))}{2} >0$, where $d$ is the metric on $Y$.



                  Now define $C={p}$, which is finite, so compact. A supremum of $d(f(x), g(x))$ with $x in C$ is then just the single value $d(f(p), g(p))$.



                  Then $f in B_C(f,varepsilon) = {h in Y^X: d(f(p), h(p)) < varepsilon}$ and $g in B_C(g,varepsilon) = {h in Y^X: d(g(p), h(p)) < varepsilon}$.



                  If $h in B_C(f, varepsilon) cap B(g, varepsilon)$ existed, then both



                  $$d(h(p), f(p)) < varepsilon text{, and } d(h(p), g(p)) < varepsilon$$



                  and so $$d(f(p), g(p)) le d(f(p), h(p)) + d(h(p), g(p)) < 2varepsilon = d(f(p), g(p)$$



                  which is a contradiction. So $B_C(f, varepsilon) cap B(g, varepsilon) = emptyset$ and $C(X,Y)$ is Hausdorff.







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                  answered Nov 16 at 8:24









                  Henno Brandsma

                  101k344107




                  101k344107






















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                      Hint: Singleton sets ${x}$ are always compact






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                        up vote
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                        Hint: Singleton sets ${x}$ are always compact






                        share|cite|improve this answer























                          up vote
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                          up vote
                          1
                          down vote









                          Hint: Singleton sets ${x}$ are always compact






                          share|cite|improve this answer












                          Hint: Singleton sets ${x}$ are always compact







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 16 at 3:10









                          bitesizebo

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                          1,39618






























                               

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