Chapter 4.SE, Problem 1SE

Consider the following properties a space may satisfy:

(1) connected

(2) path connected

(3) locally connected

(4) locally path connected

(5) compact

(6) limit point compact

(7) locally compact Hausdorff

(8) Hausdorff

(9) regular

(10) completely regular

(11) normal

(12) first-countable

(13) second-countable

(14) Lindelof

(15) has a countable dense subset

(16) locally rnetrizable

(17) metrizable

For each of the following spaces, determine (if you can) which of these properties it satisfies. (Assume the Tychonoff theorem if you need it.)

(a) $S_{\Omega }$

(b) ${\bar{S}}_{\Omega }$

(c) $S_{\Omega }\times {\bar{S}}_{\Omega }$

(d) The ordered square

(e) ${ℝ}_{\mathcal{l}}$

(f) ${ℝ}_{\mathcal{l}}^2$

(g) ${ℝ}^{\omega }$ in the product topology

(h) ${ℝ}^{\omega }$ in the uniform topology

(i) ${ℝ}^{\omega }$ in the box topology

(j) ${ℝ}^I$in the product topology, where $I=\left[0,1\right]$

(k) ${ℝ}_K$

To determine

(a)

The properties satisfied by $S_{\Omega }$ from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by $S_{\Omega }$ are given below.

Explanation of Solution

Given:

Consider the following properties a space may satisfy:

(1) connected

(2) path connected

(3) locally connected

(4) locally path connected

(5) compact

(6) limit point compact

(7) locally compact Hausdorff

(8) Hausdorff

(9) regular

(10) completely regular

(11) normal

(12) first-countable

(13) second-countable

(14) Lindelof

(15) has a countable dense subset

(16) locally rnetrizable

(17) metrizable

Approach:

Some basic results of topology for a space $X$ are given below

1. If $X$ is path connected, then it is connected.
2. If $X$ is locally path connected, then it locally connected.
3. If $X$ is compact, then it is limit point compact, locally compact and Lindelof.
4. If $X$ is compact and Hausdorff, then it is normal.
5. If $X$ is regular and Lindelof, then it is normal.
6. If $X$ is normal, then it is completely regular which implies it is regular and thus, Hausdorff.
7. If $X$ is second-countable, then it is first-countable, Lindelof and separable.
8. If $X$ is metrizable, then it is locally metrizable and normal.
9. If $X$ is Lindelof and metrizable, then it is second-countable.
10. If $X$ is separable and metrizable, then it is second-countable.
11. If $X$ is locally metrizable, then it is first-countable.
12. If a linearly arranged set $X$ has immediate successor for every non-greatest element, then it is disconnected in the order topology. In particular, this properly holds for a well-ordered set.
13. If $X$ is a completely disconnected space with at least two points, then it is disconnected.
14. A completely disconnected space is locally connected if only if it is discrete.
15. If $X\times Y$ has any other properties except for the limit point compactness, so do and $Y$.

Calculation:

Consider the space $S_{\Omega }$.

1. The space $S_{\Omega }$ consists of non-least element having no immediate predecessor. So, $S_{\Omega }$ is not connected locally and thus, not connected.
2. From above, $S_{\Omega }$ could be limit point compact but not compact.
3. Consider the set $\left[{\alpha }_0,\alpha \right]$ for ${\alpha }_0,\alpha \in S_{\alpha }$. Here, ${\alpha }_0$ is the least element and it is the neighborhood of $\alpha$ which is compact. So $S_{\Omega }$ it is locally compact.
4. $S_{\Omega }$ is normal but not Lindelof, since the open cover $\left\{\left[{\alpha }_0,\alpha \right)|\alpha \in S_{\Omega }\right\}$ does not have a countable sub cover.
5. Consider $A\subseteq S_{\Omega }$ is countable. Then, $A$ has upper bound $\alpha$ such that $A\subseteq \left[{\alpha }_0,\alpha \right)\ne S_{\Omega }$. Thus, $S_{\Omega }$ is not separable.
6. $S_{\Omega }$ is locally metrizable since has the neighborhood $\left[{\alpha }_0,\alpha \right]$ which is not regular and second countable and so, not metrizable.
7. $S_{\Omega }$ is not metrizable as it is limit point compact but not compact.

Conclusion:

Hence, the properties satisfied by $S_{\Omega }$ are given above.

To determine

(b)

The properties satisfied by ${\overline{S}}_{\Omega }$ from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by ${\overline{S}}_{\Omega }$ are given below.

Explanation of Solution

Consider the space $S_{\Omega }$.

1. Since $S_{\Omega }$ is not connected or locally connected, ${\overline{S}}_{\Omega }$ is also not connected or path connected and thereby it is not path connected or locally path connected.
2. is compact and thus, limit point compact, Lindelof and locally compact.
3. ${\overline{S}}_{\Omega }$ is Hausdorff since every order topology in ${\overline{S}}_{\Omega }$ is Hausdorff. This, it is normal and hence completely regular and regular.
4. The space is not first-countable, hence not locally metrizable and so, not metrizable. Also, it is not second-countable.
5. Let $A$ be any countable subset of ${\overline{S}}_{\Omega }$. Then, $A\cap S_{\Omega }$ has an upper bound $\alpha \in S_{\Omega }$ such that $\overline{A}\subseteq \left[{\alpha }_0,\alpha \right)\cup \left\{\Omega \right\}\ne {\overline{S}}_{\Omega }$. Thus, ${\overline{S}}_{\Omega }$ is not separable.

Conclusion:

Hence, the properties satisfied by ${\overline{S}}_{\Omega }$ are given above.

To determine

(c)

The properties satisfied by $S_{\Omega }\times {\overline{S}}_{\Omega }$ from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by $S_{\Omega }\times {\overline{S}}_{\Omega }$ are given below.

Explanation of Solution

Consider the space $S_{\Omega }\times {\overline{S}}_{\Omega }$.

1. $S_{\Omega }\times {\overline{S}}_{\Omega }$ is limit point compact.

   Since, ${\pi }_1\left(A\right)$ is countable, it has an upper bound $\alpha \in S_{\Omega }$. Here $A$ is limited in the compact space $\left[{\alpha }_0,\alpha \right]\times {\overline{S}}_{\Omega }$ and hence it has a limit point. The neighborhood of any point in $S_{\Omega }\times {\overline{S}}_{\Omega }$ has the form $\left[{\alpha }_0,\alpha \right]\times {\overline{S}}_{\Omega }$ and hence, $S_{\Omega }\times {\overline{S}}_{\Omega }$ is locally compact.

2. $S_{\Omega }\times {\overline{S}}_{\Omega }$ is regular since it is the product of completely regular spaces and hence, is resular and Hausdorff.
3. $S_{\Omega }\times {\overline{S}}_{\Omega }$ is neither normal nor compact. Also, it is neither second-countable nor Lindelof.
4. $S_{\Omega }\times {\overline{S}}_{\Omega }$ does not have any other property, since both $S_{\Omega }$ and ${\overline{S}}_{\Omega }$ do not satisfy any other properties, together.

Conclusion:

Hence, the properties satisfied by $S_{\Omega }\times {\overline{S}}_{\Omega }$ are given above.

To determine

(d)

The properties satisfied by $I_0^2$ from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by $I_0^2$ are given below.

Explanation of Solution

Consider the space $I_0^2$:

1. $I_0^2$ is connected but not path connected. The sets $x\times I,x\in I$ are path components of $I_0^2$ and since these are not open, $I_0^2$ is not locally path connected.
2. $I_0^2$ is Hausdorff and compact since all order topologies on $I_0^2$ are Hausdorff. Then, it is normal and hence, regular and completely regular.
3. $I_0^2$ is first-countable.
4. If $D$ is any dense subset of $I_0^2$, then meets $x\times \left(0,1\right)$ for $x\in I$ and hence ${\pi }_1$ maps $D$ onto $I$. So $I_0^2$ is not separable.
5. Any neighborhood of $0\times 1$ has some set $\left[a,b\right]\times I$ that is homeomorphic to $I_0^2$. So, $I_0^2$ is lindelof but not second-countable.
6. $I_0^2$ is neither metrizable nor locally metrizable.

Conclusion:

Hence, the properties satisfied by $I_0^2$ are given above.

To determine

(e)

The properties satisfied by ${ℝ}_l$ from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by ${ℝ}_l$ are given below.

Explanation of Solution

Consider the space ${ℝ}_l$.

1. ${ℝ}_l$ is not connected. For $x<y$, the separation $\left\{\left(-\infty ,y\right),\left[y,\infty \right)\right\}$ shows that they are in different components. So, ${ℝ}_l$ is neither connected locally nor connected.
2. The subset $\mathbb{Z}$ does not have a limit point and hence, ${ℝ}_l$ is not limit point compact. It is Hausdorff but not locally compact.
3. ${ℝ}_l$ is normal and so, regular and completely regular.
4. It is first-countable but not second-countable. It is separable and Lindelof but is neither metrizable nor locally metrizable.

Conclusion:

Hence, the properties satisfied by ${ℝ}_l$ are given above.

To determine

(f)

The properties satisfied by ${ℝ}_l^2$ from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by ${ℝ}_l^2$ are given below.

Explanation of Solution

Consider the space ${ℝ}_l^2$.

1. ${ℝ}_l^2$ is not normal and not limit point compact since ${\mathbb{Z}}^2$ has no limit point.
2. Since ${ℝ}_l$ is not connected or locally connected or locally compact or locally metrizable, same is true for ${ℝ}_l^2$.
3. Since ${ℝ}_l$ is regular, first-countable and separable, same is true for ${ℝ}_l^2$.

Conclusion:

Hence, the properties satisfied by ${ℝ}_l^2$ are given above.

To determine

(g)

The properties satisfied by ${ℝ}^{\omega }$ in the product topology from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by ${ℝ}^{\omega }$ in the product topology are given below.

Explanation of Solution

Consider the space ${ℝ}^{\omega }$ in the product topology.

1. ${ℝ}^{\omega }$ in the product topology is locally path connected and path connected, because it is the product of path connected and locally path connected spaces.
2. ${ℝ}^{\omega }$ is neither limit point compact nor locally compact in the product topology.
3. ${ℝ}^{\omega }$ is metrizable and second countable.

Conclusion:

Hence, the properties satisfied by ${ℝ}^{\omega }$ in the product topology are given above.

To determine

(h)

The properties satisfied by ${ℝ}^{\omega }$ in the uniform topology from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by ${ℝ}^{\omega }$ in the uniform topology are given below.

Explanation of Solution

Consider the space ${ℝ}^{\omega }$ in the uniform topology.

1. ${ℝ}^{\omega }$ in the uniform topology is not connected but is path connected and locally path connected.
2. For $r$ with $0<r<\frac{1}{2}$, consider the set $A_r$ of $a=\left(a_n\right)\in {ℝ}^{\omega }$ with $a_n=\pm r$ for all $n$. For $a\ne b\in A$, $\overline{\rho }\left(a,b\right)=2r$. If $x$ is a limit point of $A$, then $a\in A$ with $\overline{\rho }\left(x,a\right)<r$ and $a\ne x$. So, $B_{\overline{\rho }}\left(x,r\right)\cap A_r=\left\{a\right\}$ and $B_{\overline{\rho }}\left(x,\overline{\rho }\left(x,a\right)\right)\cap A_r=\varnothing$ which is a contradiction. So, ${ℝ}^{\omega }$, in the uniform topology, is neither limit point compact nor locally compact.
3. Let $D$ be any dense set in ${ℝ}^{\omega }$, in the uniform topology. For $\alpha \in A_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, select $f\left(a\right)\in D\cap B_{\overline{\rho }}\left(\alpha ,\frac{1}{2}\right)$. The $f$ is an injective map from $A_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ to $D$. So, ${ℝ}^{\omega }$ is not separable and by definition, is metrizable.

Conclusion:

Therefore, the properties satisfied by ${ℝ}^{\omega }$ in the uniform topology are given above.

To determine

(i)

The properties satisfied by ${ℝ}^{\omega }$ in the box topology from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by ${ℝ}^{\omega }$ in the box topology are given below.

Explanation of Solution

Consider the space ${ℝ}^{\omega }$ in the box topology.

1. ${ℝ}^{\omega }$ in box topology is not connected. Consider the separation $\left\{A,B\right\}$ in which $A$ is a bounded sequence set and $B$ is an unbounded sequence set. For $\alpha \in {ℝ}^{\omega }$ such that ${\alpha }_n\ne 0$ for all $n$, the function $h_{\alpha }:{ℝ}^{\omega }\to {ℝ}^{\omega }$ given by $h_{\alpha }\left(x\right)=\left({\alpha }_n{x}_n\right)$ is a homeomorphism. The neighborhood $U$ of $0$ contains a point $x$ such that $x_n\ne 0$ for all $n$.

Consider ${\alpha }_n=x_n/n$. The homeomorphism $h_{\alpha }$ takes 0 to 0 and sequence $\left(n\right)$ to $x$. So, 0 and $x$ lie on opposite sides of the separation $\left\{h_{\alpha }\left(A\right),h_{\alpha }\left(B\right)\right\}$ and thus, ${ℝ}^{\omega }$ is not locally connected.

1. Consider $D=\left\{0,1\right\}$ and take $D^{\omega }$ as the set of all the sequences of 0’s and 1’s. For $x\in {ℝ}^{\omega }$, there is a neighborhood $U_n$ of $x_n$ with $U_n\cap D\subseteq \left(x_n\right)$ for all $n$. So, $U={\displaystyle \prod U_n}$ is the box neighborhood of $x$ with $U\cap D^{\omega }\subseteq \left(x_n\right)$. Hence, $D^{\omega }$ does not have any limit points and hence ${ℝ}^{\omega }$ is not limit point compact and hence not locally compact.
2. ${ℝ}^{\omega }$ is completely regular but nothing can be said whether it is normal or not.
3. ${ℝ}^{\omega }$ is not first-countable, not Lindelof and not separable.

Conclusion:

Hence, the properties satisfied by ${ℝ}^{\omega }$ in the box topology are given above.

To determine

(j)

The properties satisfied by ${ℝ}^I$, where $I=\left[0,1\right]$, in the product topology from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by ${ℝ}^I$, where $I=\left[0,1\right]$, in the product topology are given below.

Explanation of Solution

Consider the space ${ℝ}^I$, where $I=\left[0,1\right]$ in the product topology.

1. ${ℝ}^I$ in the product topology is locally path connected and path connected because it is product of locally path connected and path connected spaces.
2. ${ℝ}^I$ is neither limit point compact, nor locally compact since no set containing a neibhorhood of origin is limit point compact.
3. The space is completely regular since it is the product of completely regular spaces and so, it is also not normal.
4. The space is separable but not first-countable.

Conclusion:

The properties satisfied by ${ℝ}^I$, where $I=\left[0,1\right]$, in the product topology are given above.

To determine

(k)

The properties satisfied by ${ℝ}_K$ from the given list.

Answer to Problem 1SE

Solution:

The properties satisfied by ${ℝ}_K$ are given below.

Explanation of Solution

Consider the space ${ℝ}_K$.

1. The open set $\left(0,\infty \right)$ which receives its typical topology, where it is not connected. Thus, $\overline{\left(0,\infty \right)}=\left(0,\infty \right)$ is connected.
2. The open set $ℝ-K$ also receives its typical topology having many numbers of consequences. One constituent of $ℝ-K$ is $(-\infty ,0]$ which is not open in ${ℝ}_K$ and hence, ${ℝ}_K$ is not locally connected but as union of subsets $(-\infty ,0]$ and $\left(0,\infty \right)$, it is connected. Also, it is locally metrizable for metrizable union of subsets $\left(0,\infty \right)$ and $ℝ-K$.
3. A map from ${ℝ}_K$ to $ℝ$ is the identity map which is continuous. So, ${ℝ}_K$ has a connected subspace in the typical topology. Conversely, $f:I\to {ℝ}_K$ is a path having $f\left(0\right)\le 0$ and $f\left(1\right)>0$. Suppose $f\left(s\right)<0$, for $t>s$ with $f\left(t\right)=0$ because $f\left(\left[s,1\right]\right)$ is connected. Since $f^{-1}\left(\left(-\infty ,0\right]\right)$ is closed and has the biggest element $s_0<1$, $f\left((s_0,1]\right)\subseteq \left(0,\infty \right)$. The neighborhood of 0 is $ℝ-K$ which has $s>s_0$ with $f\left([s_0,s]\right)\subseteq ℝ-K$, since $f\left([s_0,s]\right)$ is connected, so $\left[0,f\left(s\right)\right]\subseteq f\left(\left[s_0,s\right]\right)$ is a contradiction. So, ${ℝ}_K$ is not path connected.
4. Also, ${ℝ}_K$ is Hausdorff but not regular and is not limit point compact.
5. ${ℝ}_K$ is second-countable because it has a basis which has all sets $\left(a,b\right)$ and $\left(a,b\right)-K$, where $a$ and $b$ are rationals.

Conclusion:

Hence, the properties satisfied by ${ℝ}_K$ are given above.