Let k be an infinite field.Let V be a vector space over k and W 1 , . . . , W r proper subspaces of V.Show that ⋃ i = 1 r W i ≠ V .I tried the following:for all j ∈ { 1 , . . . , r }, I take w j ∈ W j such that w j ∉ W i whenever j ≠ i, so I know that w 1 + ⋯ + w r ∈ V. If w 1 + ⋯ + w r ∈ ⋃ i = 1 r W i , then there is l ∈ { 1 , . . . , r } such that w 1 + ⋯ + w r ∈ W l . I don't find because w 1 + ⋯ + w r ∈ W l is absurd.Is this correct reasoning, or is there other way for me to prove this?

Question

Let k be an infinite field.Let V be a vector space over k and       W    1    ,  .  .  .  ,      W    r   proper subspaces of V.Show that       ⋃          i      =      1        r        W    i    ≠  V  .I tried the following:for all   j  ∈  {  1  ,  .  .  .  ,  r  }, I take       w    j    ∈      W    j   such that       w    j    ∉      W    i   whenever   j  ≠  i, so I know that       w    1    +  ⋯  +      w    r    ∈  V. If       w    1    +  ⋯  +      w    r    ∈      ⋃          i      =      1        r        W    i  , then there is   l  ∈  {  1  ,  .  .  .  ,  r  } such that       w    1    +  ⋯  +      w    r    ∈      W    l  . I don&#039;t find because       w    1    +  ⋯  +      w    r    ∈      W    l   is absurd.Is this correct reasoning, or is there other way for me to prove this?

Caylee Davenport · Accepted Answer

The assertion seems to be false. Take   k  =            F        2   and   V  =  k  ⊕  k  =  {  (  0  ,  0  )  ,  (  1  ,  0  )  ,  (  0  ,  1  )  ,  (  1  ,  1  )  }. Now you can take r=3 and the following proper subspaces of       W    1    =  {  (  0  ,  0  )  ,  (  1  ,  0  )  },       W    2    =  {  (  0  ,  0  )  ,  (  1  ,  1  )  }, and       W    3    =  {  (  0  ,  0  )  ,  (  1  ,  1  )  }