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Unit 5: The Subspace Topology
Let G be arbitrary, then G T, by (1). Notes
We can write G = H Y for some T-open set H.
Now, G = H Y, G T H Y T
Again H Y T, H T and Y X Y T
Conversely, let (Y, ) (X, T) and let Y T for any G G T
G A T s.t. G = A Y
Again A T, Y T A Y T G T
Finally, any G G T.
5.1.2 Basis for the Subspace Topology
Example 8: Consider the subset Y = [0, 1] of the real line , in the subspace topology. The
subspace topology has as basis all sets of the form (a, b) Y, where (a, b) is an open interval in
, such a set is of one of the following types:
ì (a,b) if a and b are in Y,
ï
ï [0,b) if only b is in Y,
(a, b) Y = í
ï (a,1] if only a is in Y,
ï Y or if neither a nor b is in Y.
î
By definition, each of these sets is open in Y. But sets of the second and third types are not open
in the larger space .
Note that these sets form a basis for the order topology on Y. Thus, we see that in the case of the
set Y = [0, 1], its subspace topology (as a subspace of ) and its order topology are the same.
Example 9: Let Y be the subset [0, 1) {2} of . In the subspace topology on Y the one-
5
3
point set {2} is open, because it is the intersection of the open set ( , ) with Y. But in the order
2 2
topology on Y, the set {2} is not open. Any basis element for the order topology on Y that
contains 2 is of the form
{x x Y and a < x 2}
for some a Y; such a set necessarily contains points of Y less than 2.
Lemma 1: If is a basis for the topology of X, then the collection = {B Y : B } is a basis for
Y
the subspace topology on Y.
Proof: Given -open in X and given y Y, we can choose an element B of such that
y B . Then y B Y Y
Now as we know
“If X is a topological space and is a collection of open sets of X such that for each open set of
X and each x in , there is an element c of such that x . The is a basis for the topology
of X.”
Thus, we can say that is a basis for the subspace topology on Y.
Y
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