Worksheet 2

Question 1 — Basis for Subspace Topology

Let be a topological space and let be a basis for .

Suppose and let be the subspace topology on :

Define

Prove that is a basis for .

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Proof

We verify the basis conditions.

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1. Covering Property

Let .

Since is a basis for , there exists such that

Then

Thus:

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2. Intersection Property

Let

where

Then

Since is a basis, there exists such that

Hence

Therefore is a basis.

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Question 2 — Subspace Topologies of Lines in

Let have the standard topology .

Let have the standard topology .

Define maps:

(horizontal axis)

(vertical axis)

(diagonal line)

Let

• be the subspace topology on
• be the subspace topology on
• be the subspace topology on

Prove:

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Proof for

Let

be open in the subspace topology.

Then

for some open set .

Let

Since is open, there exists such that

Intersecting with the x-axis gives:

Thus open sets correspond exactly to open intervals in .

Hence:

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Proof for

Similarly,

where is open in .

If

then some rectangle

lies in .

Intersecting with the y-axis gives:

Hence:

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Proof for

Let

where is open in .

If

then some rectangle

lies in .

Intersecting with the diagonal gives:

This corresponds to an open interval in .

Hence:

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Conclusion

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Question 3 — Subspace of Product Space

Let and be topological spaces.

Let:

with subspace topologies:

Let:

with product topology .

Consider:

with subspace topology:

Prove:

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Proof

A basis for is:

Subspace topology basis:

But:

Since:

these sets form a basis for .

Thus both topologies have the same basis.

Therefore: