Midsem

Question 1

Let

be the collection of all semi-open intervals in .

(a) Prove that is a basis for a topology on .

Solution

To show that is a basis, we verify the two basis axioms.

Axiom 1: Covering

For every , the interval

belongs to and contains . Hence every point of lies in some element of .

Axiom 2: Refinement under intersections

Let

and let . Then

So

is a semi-open interval containing and contained in the intersection. If the intersection is empty, there is nothing to prove.

Thus is a basis for a topology on .

So

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(b) Compare the topologies and .

Here denotes the topology generated by , and denotes the standard topology on .

Solution

The topology is strictly finer than the standard topology.

Indeed, every standard open interval can be written as

and each belongs to . Therefore every standard open set is open in .

So

The inclusion is strict because belongs to but is not open in the standard topology.

Therefore

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Let

(c) Is a basis for ? Justify.

Solution

No, is not a basis for .

To be a basis for , every set in would have to be expressible as a union of elements of .

Consider the basic open set

If were a basis, then for the point there would exist a rational semi-open interval

such that

But if , then . Since and is irrational, this forces . Then the interval contains points smaller than , so it cannot be contained in .

This contradiction shows that such a basis does not exist.

Hence

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(d) Is connected? Justify.

Solution

The topology generated by is the lower limit topology, also called the Sorgenfrey topology.

Consider the two subsets

Both are open in :

• for , we have

a union of basic open sets;

• for , we have

again a union of basic open sets.

They are nonempty, disjoint, and their union is all of :

Thus is separated into two disjoint nonempty open sets, so it is disconnected.

Therefore

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Question 2

Let and be topological spaces and . Suppose .

(a) When is said to be continuous?

Solution

The map is continuous if and only if for every open set , the preimage is open in ; that is,

So

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(b) If is continuous, is the restriction continuous? Justify.

Solution

Yes, the restriction is continuous.

Let be open. Then

Since is continuous, is open in , and hence is open in the subspace topology on .

Therefore is continuous.

So

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Suppose with the standard topology. Let be defined by

(c) Is continuous? Justify.

Solution

Yes. On the subspace , the function is identically .

Indeed, for every we have . A constant map is continuous.

Thus

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(d) Is continuous? Justify.

Solution

Yes. On the subspace , the function is identically .

So it is a constant map, hence continuous.

Therefore

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(e) Is continuous on ? Justify.

Solution

No. The function has a jump discontinuity at .

Indeed,

so the left and right limits are different. Therefore is not continuous at .

Hence

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(f) Is continuous? Justify.

Solution

Yes. On , the function .

The identity function on a subspace of is continuous, so the restriction is continuous.

Thus

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(g) Is continuous? Justify.

Solution

Yes. On , the function .

Since polynomial functions are continuous, the restriction is continuous.

Therefore

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(h) Is continuous on ? Justify.

Solution

Yes. We check continuity at the only potentially problematic point, .

For , , so

For , , so

Also,

Thus the left limit, right limit, and function value agree at . Since is continuous on each side of and at , it is continuous on all of .

Therefore

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Question 3

Consider equipped with the Euclidean metric and the discrete metric , given by

Let

be the identity map . Verify whether is a homeomorphism.

Solution

We check whether is continuous and whether its inverse is continuous.

Step 1: Continuity of

Because the domain has the discrete metric, every subset of is open in . Therefore the preimage under of any open set in is open.

Hence is continuous.

Step 2: Continuity of the inverse map

The inverse map is again the identity,

To be continuous, the preimage of every open set in the discrete topology would have to be open in the Euclidean topology.

But in the discrete topology, every singleton is open. Its preimage under is just , and a singleton is not open in the Euclidean topology.

Thus is not continuous.

Therefore is not a homeomorphism.

So the conclusion is

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Question 4

Define the spaces as

and

Prove the following statements.

(a) is path connected.

Solution

The set consists of the horizontal segment together with the vertical segments for .

To show path connectedness, we exhibit paths joining any point of to the point .

Take any point .

• If , then the straight line path

  lies entirely in and joins to .

• If for some and , then first move vertically to and then horizontally to . This is given by the path

  where .

This path lies in because the first half stays on the vertical segment and the second half stays on the horizontal segment .

Thus every point of can be joined by a path to , so any two points of can be joined by concatenating their paths to .

Therefore is path connected.

So

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(b) is connected.

Solution

The space is path connected by part (a), hence connected.

Now observe that every point of is a limit point of in the subspace . Indeed:

• points already in are obviously in the closure of ;
• if , then for any neighborhood of in , points lie in that neighborhood for all sufficiently large .

Therefore is dense in , so

The closure of a connected set is connected. Since is connected, its closure in is connected. Hence is connected.

Therefore

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(c) is not path connected.

Solution

We prove that there is no path in from the point to any point of the horizontal base .

Suppose, for contradiction, that there exists a path

such that and .

Write , where are continuous.

Let

Then for all , and by continuity.

For , the point lies in the part of with positive second coordinate. Therefore

Since is connected and , the image is a connected subset of contained in the set

But any connected subset of this set is a singleton, because it contains no interval of positive length.

Hence is constant on . Since , we get

By continuity, as well. Therefore

which is impossible because .

This contradiction shows that no such path exists.

Therefore is not path connected.

So

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(d) is totally disconnected.

Solution

First identify the set. The intersection with the horizontal line is

Consider the projection onto the first coordinate,

This is a homeomorphism onto the subset

Now let be connected. Since the inclusion map is continuous, is connected as a subset of . But the connected subsets of are intervals, and contains no nontrivial interval. Hence every connected subset of is a singleton.

Therefore the connected components of , and hence of , are singletons.

Thus the space is totally disconnected.

Hence

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Question 5

Let be a metric space and . For any , define

(a) Prove that

Solution

Fix . For any , the triangle inequality gives

Taking the infimum over all on the right-hand side yields

Interchanging the roles of and gives

Combining the two inequalities, we obtain

Thus

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Fix and define

(b) Is the set open or closed in ? Justify.

Solution

The function is continuous by part (a), because it is -Lipschitz:

Now

Since is closed in , its preimage under a continuous map is closed in .

Therefore

It need not be open in general.

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(c) Find .

Solution

As , the sets increase:

Hence the natural set-theoretic limit is the union over all positive :

Now if and only if for some , which is equivalent to

Therefore

Indeed, exactly when lies in the closure of .

So

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Midsem Question Paper