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Homework Statement
Consider a metric space (X,d) with subsets A and B of X, where A and B have non-zero intersection. Show that diam(A[itex]\bigcup[/itex]B) [itex]\leq[/itex] diam(A) + diam(B)
Homework Equations
The Attempt at a Solution
A hint would be very much appreciated. Let x[itex]\in[/itex]A, y[itex]\in[/itex]B, z[itex]\in[/itex]A[itex]\bigcup[/itex]B
diam(A[itex]\bigcup[/itex]B) = sup[d(x,y)] for every x,y.
So d(x,y) [itex]\leq[/itex] diam(A[itex]\bigcup[/itex]B)
Either z[itex]\in[/itex]A, or z[itex]\in[/itex]B, or z in both.
Consider firstly z[itex]\in[/itex]A.
d(x,z) [itex]\leq[/itex] diam(A)
If z[itex]\in[/itex]B,
d(z,y) [itex]\leq[/itex] diam(B)
Therefore, d(x,z) + d(z,y) [itex]\leq[/itex] diam(A) + diam(B)
But d(x,y) [itex]\leq[/itex] d(x,z) + d(z,y), hence
d(x,y) [itex]\leq[/itex] diam(A) + diam(B)As far as I've gotten...