Suppose X and Y are subsets of R. Define the set X+Y as follows:?

Question

Suppose X and Y are subsets of R. Define the set X+Y as follows:
X +Y={ z of R I z for some x of X and y of Y}.
Suppose in addition that X and Y are non-empty open set. Prove is Open.

Answer 1

Since R is a metric space, and X and Y are open in R, for every a in (X+Y) there exists d1 and d2 such that B(a,d1)={x in X: |x-a|

Answer 2

to people that're answering in undemanding words by way of technique of asserting that homosexuality is incorrect, the XY gene is male, yet each so often an XY guy or woman could be born with woman genetalia. women folk are XX, yet each so often an XX guy or woman could be born with male genetalia. The XY / XX think of of is why the guy determines the gender of the toddler. All eggs carry in undemanding words an X chromosomes, yet sperm carry the two an X or a Y. So the question is how is homosexuality desperate? If an outwardly male appearing guy or woman falls in love with a guy, yet he's genetically woman, is it nonetheless homosexualaity? stable question. Edit: Kudos to Renee. That substitute top right into a severe extreme high quality answer.

Answer 3

Incompletely asked question. However, if X is open, then
X+Y is the union of X+y as y ranges over Y, so it is a union of open sets, so it is open. Similarly if Y is open. You don't need them both to be open to get the conclusion though. This proof works in any topological group, BTW.