I'm doing this exercise in the topology of Munkres (~ paraphrasing). I have two motivations for reading this book: 1, the topology is really pretty, but the main one, 2, is that I'm not yet an undergraduate student; therefore, I have never taken any kind of "evidence-based" course and would like to learn how to write legible tests. Appreciate:

- A verification of my test.
- Any advice on how to make my test more readable.

Remember that $ S _ { Omega} $ Denotes the minimum set ordered, uncountable. Let L denote the ordered set $ S _ { Omega} times[01)$[01)$[01)$[01)$ in the order of the dictionary, with its smallest element removed. The set L is a classic example in topology called **Long line**.

Theorem: the long line is connected to the route and locally homeomorphic to $ mathbb {R} $, but it can not be embedded in $ mathbb {R} $.

The structure of the test is given by Munkres.

(a) Let X be an ordered set; leave $ a <b <c $ Be X points. Show that $[ac)$[ac)$[ac)$[ac)$ has the order type of $[01)$[01)$[01)$[01)$ Yes, both $[ab)$[ab)$[ab)$[ab)$ Y $[bc)$[bc)$[bc)$[bc)$ have the order type of $[01)$[01)$[01)$[01)$.

Assume $[ab)$[ab)$[ab)$[ab)$ Y $[bc)$[bc)$[bc)$[bc)$ have the order type of $[01)$[01)$[01)$[01)$. Then there are order isomorphisms. $ f:[ab)rightarrow[0frac{1}{2})$[ab)rightarrow[0frac{1}{2})$[ab)rightarrow[0frac{1}{2})$[ab)rightarrow[0frac{1}{2})$ Y $ g:[bc)rightarrow[0frac{1}{2})$[bc)rightarrow[0frac{1}{2})$[bc)rightarrow[0frac{1}{2})$[bc)rightarrow[0frac{1}{2})$, why $[01)$[01)$[01)$[01)$ has the same kind of order that $[01/2)$[01/2)$[01/2)$[01/2)$. Define $ h:[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$ as follows:

$$ h (x) = begin {cases}

f (x) text {if} x in[a, b) \

g (x) text {if} x in[BC)

end {cases} $$

$ h $ it's an order-isomorphism, like $ p <q $ it implies $ h (p) <h (q) $.

Conversely, it assumes $[ac)$[ac)$[ac)$[ac)$ has the order type of $[01)$[01)$[01)$[01)$. Then there is an order-isomorphism. $ f:[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$. Restricting the domain to $[ab)$[ab)$[ab)$[ab)$ gives a new order-isomorphism $ g:[ab)rightarrow[0d)$[ab)rightarrow[0d)$[ab)rightarrow[0d)$[ab)rightarrow[0d)$, where $ d in (0.1) $. $[0d)$[0d)$[0d)$[0d)$ is of the order type of $[01)$[01)$[01)$[01)$; multiplying by $ 1 / d $ It is an order of isomorphism. Similarly, restricting the domain of $ f $ to $[bc)$[bc)$[bc)$[bc)$ gives order-isomorphism $ h:[bc)rightarrow[d1)$[bc)rightarrow[d1)$[bc)rightarrow[d1)$[bc)rightarrow[d1)$, where $ d in (0.1) $.

(b) Let X be an ordered set. Leave $ x_0 <x_1 <… $ Be a growing sequence of points of X; suppose $ b = sup {x_i } $. Show that $[x_0b)$[x_0b)$[x_0b)$[x_0b)$ has the order type of $[01)$[01)$[01)$[01)$ if each interval $[x_ix_i+1)$[x_ix_i+1)$[x_ix_i+1)$[x_ix_i+1)$ has the order type of $[01)$[01)$[01)$[01)$.

This is simple, having already tried. $ (a) $. Begin with $ i = 1 $: $[x_0x_1)[x_1b)$[x_0x_1)[x_1b)$[x_0x_1)[x_1b)$[x_0x_1)[x_1b)$ It is a partition of a set in the same way that we saw in $ (a) $; therefore, each must have the same type of order that $[01)$[01)$[01)$[01)$. We proceed by induction, considering now the whole. $[x_1b)$[x_1b)$[x_1b)$[x_1b)$ and the sequence $ x_1 <x_2 <… $ The test is the same.

(c) Leave $ a_0 $ denotes the smallest element of $ S _ { Omega} $. For each element $ a $ of $ S _ { Omega} $ different from $ a_0 $, shows that the interval $[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$ of $ L $ has the order type of $[01)$[01)$[01)$[01)$.

The[[*Innuendo*: Proceed by transfinite induction. Either has an immediate predecessor in $ S _ { Omega} $, or there is a growing sequence $ a_i $ in $ S _ { Omega} $ with $ a = sup {a_i } $]

**How can I order this ?:** $[a_0a)$[a_0a)$[a_0a)$[a_0a)$ is accountable by the definition of $ S _ { Omega} $. Following the trail: yes $ a $ it has an immediate predecessor, $ b $, so $[btimes0atimes0)$[btimes0atimes0)$[btimes0atimes0)$[btimes0atimes0)$ has the order type of $[01)$[01)$[01)$[01)$ (All points in this set are of the form $ b times x $, where $ x en[01)$[01)$[01)$[01)$).

By induction backwards. $ a_0 $ any finite section of $ S _ { Omega} $of which $ a $ is the largest element, is the isomorphic order for $[01)$[01)$[01)$[01)$.

Yes $ a $ is not the largest element of a finite section of $ S _ { Omega} $, then there is a growing sequence $ a_i $ in $ S _ { Omega} $ with $ a = sup {a_i } $, this sequence is exactly every element of $ S _ { Omega} $ strictly less than $ a $, in order (which must be countable). Every $ a_ {i + 1} $ In this sequence has an immediate predecessor, namely: $ a_1 $, then each $[a_{i}times0a_{i+1}times0)$[a_{i}times0a_{i+1}times0)$[a_{i}times0a_{i+1}times0)$[a_{i}times0a_{i+1}times0)$ must be isomorphic order to $[01)$[01)$[01)$[01)$ for the previous argument. Then for $ (b) $, then you must $[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$.

(d) Show that $ L $ The road is connected.

Extensible $ (c) $ to include sets of the form $[ab)abenL$[ab)abinL$[ab)abenL$[ab)abinL$ It's trivial – just add / subtract the sets of the ends – I'll assume this. This implies that $[a,b]$ has the same kind of order that $[0,1]$. So between two points $ a, b in L $, there is an order-isomorphism $ f:[0,1] correct arrow [a,b]$; which is necessarily a homeomorphism and therefore a continuous map.

**Is that last step valid? How can I make that clearer?**

(e) Show that each point of $ L $ It has a homeomorphic neighborhood with an open interval in. $ mathbb {R} $.

Leave $ x $ be our point, assume $ x neq a_0 $. Then there are some points, $ a, b in L $, such that $ a <x <b $. By $ (d) $, $ L $ The path is connected, so there is some homeomorphism. $ f:[c,d] correct arrow [a,b]$ such that $ f (c) = a $, $ f (d) = (b) $. Then, restricting the range of $ f $ to $ (c, d) $ we get a homeomorphism $ f & # 39; 🙁 c, d) rightarrow (a, b) $.

And finally…

(f) Show that L can not be embedded in $ mathbb {R} $or indeed in $ mathbb {R} ^ n $ For any $ n $.

The[[*Innuendo*: any subspace of $ mathbb {R} ^ n $ It has an accounting basis for its topology.

Suppose there was an imbedding $ f: L rightarrow mathbb {R} ^ n $, therefore a homeomorphism $ f & # 39 ;: L rightarrow Y $, $ Y subset mathbb {R} ^ n $ obtained by restricting the range of $ f $. $ Y $ It must have an accounting basis because $ mathbb {R} ^ n $ is metrizable, so $ L $ It must also have an accounting base for its topology, being homeomorphic with $ Y $.

Assume $ L $ I had an accounting basis, call this base $ U_ {i} $ where $ i in mathbb {Z} ^ {+} $. Each open set of $ L $ It must contain, in its entirety, at least one. $ U_i $. Leave $ x _ { alpha} $ Be the collection of all the points of the form. $ y times 0 $ where $ y in S $ {Omega} $. $ x _ { alpha} $ It is uncountable, being indexed by $ S _ { Omega} $. Then for (e), there is a countless collection $ V _ { alpha} $ of neighborhoods around $ x _ { alpha} $. Keep in mind that we can choose, with more force, $ V _ { alpha} $ in such a way that it is paired-disunion.

**That last step feels neglected; Any suggestions?**

Then each contains some $ U_ {i} $; So, there is some injective function. $ f: S _ { Omega} rightarrow mathbb {Z} ^ {+} $ defined by $ f ( alpha) = i $ as long as $ U_i subset V _ { alpha} $. This contradicts that $ S _ { Omega} $ it is countless

$ therefore $