For arbitrary subspaces U, V and W of a finite dimensional vector spac...
Proof for each option:
(a) U∩(V ∩ W) ⊆ U∩V ∩ U∩W:
Let x be an arbitrary element in U∩(V ∩ W). This means that x belongs to both U and (V ∩ W). Therefore, x belongs to U and x belongs to V ∩ W. From this, we can conclude that x belongs to U and x belongs to V and x belongs to W. Hence, x belongs to U∩V and x belongs to U∩W. Therefore, x belongs to U∩V ∩ U∩W. Since x was arbitrary, we have shown that U∩(V ∩ W) ⊆ U∩V ∩ U∩W.
(b) U∩(V ∩ W) ⊇ U∩V ∩ U∩W:
Let x be an arbitrary element in U∩V ∩ U∩W. This means that x belongs to both U∩V and U∩W. Therefore, x belongs to U and x belongs to V, and x belongs to U and x belongs to W. From this, we can conclude that x belongs to both U and (V ∩ W). Hence, x belongs to U∩(V ∩ W). Therefore, x belongs to U∩(V ∩ W). Since x was arbitrary, we have shown that U∩(V ∩ W) ⊇ U∩V ∩ U∩W.
(c) (U∩V) ∩ (U∩W) ⊆ (U∩W)∩(V∩W):
Let x be an arbitrary element in (U∩V) ∩ (U∩W). This means that x belongs to both U∩V and U∩W. Therefore, x belongs to U and x belongs to V, and x belongs to U and x belongs to W. From this, we can conclude that x belongs to (U∩W) and x belongs to (V∩W). Hence, x belongs to (U∩W)∩(V∩W). Therefore, x belongs to (U∩W)∩(V∩W). Since x was arbitrary, we have shown that (U∩V) ∩ (U∩W) ⊆ (U∩W)∩(V∩W).
(d) (U∩V) ∩ W ⊆ (U∩W) ∩ (V∩W):
Let x be an arbitrary element in (U∩V) ∩ W. This means that x belongs to both U∩V and W. Therefore, x belongs to U and x belongs to V, and x belongs to W. From this, we can conclude that x belongs to (U∩W) and x belongs to (V∩W). Hence, x belongs to (U∩W) ∩ (V∩W). Therefore, x belongs to (U∩W) ∩ (V∩W). Since x was arbitrary, we have shown that (U∩V) ∩ W ⊆
For arbitrary subspaces U, V and W of a finite dimensional vector spac...
I think obtion b is correct