# Tutorial Problems I: Topology: Hocking and Young

Reference: Topoology by Hocking and Young, Dover Publications, Inc., NY. Available in Amazon India.

Exercises 1-1:

Show that if S is a set with the discrete topology and $f: S \rightarrow T$ is any transformation of S into a topologized set T, then f is continuous.

Solution 1-1:

Definition A: The set S has a topology (or is topologized) provided that, for every point p in S and every subset X of S, the question : “is p a limit point of X?” can be answered.

Definition B: A topology is said to be a discrete topology when we assume that for no point p in S, and every subset X of S: the answer to the question: “is p a limit point of X?” is NO.

Definition C: A transformation $f: S \rightarrow T$ is continuous provided that if p is a limit point of a subset X of S, then f(p) is a limit point or a point of f(X).

So, the claim is vacuously true. QED.

Exercises 1-2:

A real-valued function $y=f(x)$ defined on an interval [a,b] is continuous provided that if $a \leq x_{0} \leq b$ and $\epsilon >0$, then there is a number $\delta>0$ such that if $|x-x_{0}|<\delta$, where $x \in [a,b]$, then $|f(x)-f(x_{0})|<\epsilon$. Show that this is equivalent to our definition, using definition 1-1.

Solution 1-2:

Definition 1-1: The real number p is a limit point of a set X of real numbers provided that for every positive number $\epsilon$, there is an element x of the set X such that $0<|p-x|<\epsilon$.

Definition C: A transformation $f: S \rightarrow T$ is continuous provided that if p is a limit point of a subset X of S, then f(p) is a limit point or a point of f(X).

Part 1: Let us assume that given function f is continuous as per definition given just above.

Then, as p is a limit point of X: it means: For any $\delta>0$, there exists a real number p such that there is an element $x \in X$ such that $|p-x|<\delta$..

So, also, by definition C, f(p) is a limit point or a point of f(X); this means the following: if f(p) is a point of f(X), there exists some $x_{0} \in X$ such that $f(x_{0} \in f(X)$, and so quite clearly in this case $p=x_{0}$ so that $|p-x_{0}|=|x_{0}-x_{0}|<\delta$, as $\delta$ is positive.

On the other hand, if f(p) is a limit point of f(X), as per the above definition of continuity, then also for any $\epsilon>0$, there exists a point $y \in f(X)$ such that $|y-f(p)|<\epsilon$. So, in this case also the claim is true.

We have proved Part 1. QED.

Now, part II: We assume the definition of continuity given in the problem statement is true. From here, we got to prove definition C as the basic definition given by the authors.

But this is quite obvious as in this case $p=x_{0}$.

We have proved Part II. QED.

Thus, the two definitions are equivalent.

Cheers,

Nalin Pithwa

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