Subbasis proof check
Question: let S be subsets of reals that only have two elements. show that it's a subbasis for it.
is my solution correct?
Let T={{0,x}|x\in R\{0}}\subseteq S. Then we want to show \bigcup T=R
Let y \in \bigcup T. Since union of sets of reals is real, y\in R
Let y\in R. Since y\in {0,y}, then y\in \bigcup T. If y=0, since y\in {0,y}, y in bigcup T
So \bigcup T=R and since a subset of S covers R, S covers R
You’re on the right track—showing that the union covers R is exactly the condition for being a subbasis
(for the topology it generates). But there’s a small glitch in your proof at y = 0.
Clean proof
Let X = R and
S = {{a, b} : a, b ∈ R, a =
 b}.
To show S is a subbasis on X , it suffices to show ⋃ S = X .
Fix y ∈ R.
If y =
 0, then {0, y} ∈ S and y ∈ {0, y}.
If y = 0, pick any x =
 0 (say x = 1). Then {0, 1} ∈ S and 0 ∈ {0, 1}.
Hence every y ∈ R lies in some member of S , so ⋃ S = R. Therefore S is a subbasis (for the topology
it generates).
What was off in your writeup
You defined T = {{0, x} : x ∈ R ∖ {0}} ⊆ S and argued ⋃ T = R. That’s fine, but your line “If y = 0,
since y ∈ {0, y}, y ∈ ⋃ T ” is not correct because {0, 0} = {0} ∈
/ T . Instead, handle y = 0 by noting
0 ∈ {0, 1} ∈ T . With that fix, your approach works: since ⋃ T = R, certainly ⋃ S = R as well.
(Bonus) What topology does it generate?
Finite intersections of sets in S include singletons, e.g.
{a, b} ∩ {a, c} = {a}.
Unions of such finite intersections are arbitrary unions of singletons, i.e. all subsets of R. So the
topology generated by S is the discrete topology.
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to show that it generates the discrete topology, i did this:
let x\in R\{0,1}. Then {0,x}\cap {0,x}={x}. So intersection of subbasis elements is open
{0,1}\cap {1,2}={1} and {0,1}\cap {0,2}={0} are also open. so all singletons are open
Good instinct, but there are a couple fixes:
1. {0, x} ∩ {0, x} = {0, x}, not {x}.
To get a singleton you need two different 2-element sets that share exactly one point.
2. A clean, uniform argument for any x ∈ R:
Pick y =
 x and z =
 x, y (possible since R has at least three distinct points).
Then
{x, y} ∈ S,
{x, z} ∈ S,
and {x, y} ∩ {x, z} = {x}.
So every singleton {x} is a finite intersection of members of S .
3. Since singletons are in the family of finite intersections (i.e., in the generated base), any set A ⊆ R is
a union of singletons:
A = ⋃ {x}.
​
x∈A
Thus every subset of R is open in the topology generated by S . That topology is the discrete topology.
(Your special cases {0, 1} ∩ {1, 2} = {1} and {0, 1} ∩ {0, 2} = {0} are perfect examples for 1 and 0;
the step that needed correction was {0, x} ∩ {0, x}.)
i meant {0,x}\cap {1,x}={x}. would that fix it? think
Yep—that fixes it (with tiny case-handling).
For any x ∈ R ∖ {0, 1}:
{0, x} ∩ {1, x} = {x}.
For x = 0:
{0, 1} ∩ {0, 2} = {0}.
For x = 1:
{0, 1} ∩ {1, 2} = {1}.
So every singleton {x} is a finite intersection of subbasis elements. Finite intersections of a subbasis
form a base; unions of base elements give all opens. Since any A ⊆ R is ⋃x∈A {x}, all subsets are open
​
—the generated topology is discrete.
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