Determine whether the group of real numbers with addition and a connected topology can have a proper connected subgroup | Step-by-Step Solution

What We're Solving:

We need to determine whether it's possible for the group of real numbers under addition (ℝ, +) to have a connected topology such that there exists a proper subgroup that is also connected. A "proper" subgroup means a subgroup that isn't the trivial subgroup {0} or the whole group ℝ itself.

The Approach:

This is a beautiful problem that connects topology with algebra! Our strategy will be to use a fundamental theorem about connected subgroups of topological groups. We'll examine what it means for a subgroup to be connected and then apply some powerful results from topological group theory.

The key insight is to use the fact that in a connected topological group, any connected subgroup must have a very specific relationship with the whole group.

Step-by-Step Solution:

Step 1: Understand what we're working with

• We have (ℝ, +) as our group
• We're given that ℝ has some connected topology (not necessarily the usual topology!)
• We want to know if there can be a proper subgroup H where H ≠ {0} and H ≠ ℝ, such that H is connected in the subspace topology

Step 2: Recall the key theorem There's a fundamental result in topological group theory: In a connected topological group G, if H is a connected subgroup, then H is dense in G.

This means that the closure of H equals the entire group G.

Step 3: Apply this to our situation Since we're assuming ℝ has a connected topology, any connected subgroup H of ℝ must be dense in ℝ.

Step 4: Use properties of ℝ as an additive group Here's the crucial insight: What are the possible subgroups of (ℝ, +)?

• The trivial subgroup {0}
• Subgroups of the form aℤ = {..., -2a, -a, 0, a, 2a, ...} for some a > 0
• Dense subgroups like ℚ (rationals)
• The whole group ℝ

Step 5: Check which subgroups can be connected

• {0} is connected (it's a single point) but it's trivial, not proper
• Any subgroup aℤ is discrete in any reasonable topology on ℝ, so it cannot be connected (unless it's just {0})
• ℝ itself is connected by assumption, but it's not proper
• Any other subgroup would need to be dense (by our theorem), but the only dense subgroup that could potentially be closed is ℝ itself

Step 6: Reach the conclusion Since any connected subgroup must be dense, and the only subgroups of ℝ that can be both dense and closed are {0} and ℝ itself, there cannot be a proper connected subgroup.

The Answer:

The statement is TRUE. The group of real numbers under addition, when equipped with any connected topology, cannot contain a proper connected subgroup.

The proof relies on the fundamental theorem that connected subgroups of connected topological groups are dense, combined with the specific structure of subgroups of (ℝ, +).

Memory Tip:

Remember this principle: "Connected subgroups spread out!" In a connected topological group, any connected subgroup must be dense - it gets arbitrarily close to every point in the group. Since ℝ has a very rigid subgroup structure, this density property forces the subgroup to be either trivial or the whole space.

This is a wonderful example of how algebraic structure (the subgroups of ℝ) constrains topological possibilities (what can be connected)!