2
$\begingroup$

I want to demonstrate the topologic definition of continuity, using the classical definition with epsilon's and delta's.

So we have that:

Classical definition - If the function $f:X \rightarrow Y$ is continuos in $x_o$ then: $\forall \varepsilon >0, \forall x\in X, \exists \delta>0: |x-x_0|<\delta \Rightarrow |f(x)-f(x_o)|<\varepsilon $

Topological definition - $f:X \rightarrow Y$ is continuous if for each open subset $V\subset Y$, $f^{−1}(V)$ is an open subset of $X$.

How could I do this? I'm a beginner in Mathematical Physics. I tried drawing graphics and the first definition seems easy to understand but I can't even visualize the second.

$\endgroup$
  • $\begingroup$ What kind of spaces are $X$ and $Y$? $\endgroup$ – principal-ideal-domain Sep 15 '15 at 17:43
5
$\begingroup$

A nice property of metric spaces is the fact that we may characterize the notion of continuity of a mapping $f$ in four different, yet equivalent ways.

Theorem 1: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Then, the following are equivalent statements:

$f$ is continuous at $a \in X$.

i) $\epsilon-\delta$ requirement.

ii) Suppose $U \subset Y$ is open in $Y$. Then, $f^{-1}(U)$ is open in $ X$.

iii) Suppose $E \subset Y$ is closed in $Y$. Then, $f^{-1}(E)$ is closed in $X$.

iv) If $\{x_n\}_{n \in \mathbb{N}} \to x$, then $\{f(x_n)\}_{n \in \mathbb{N}} \to f(x)$.

Continuity as a Motivation for Topological Spaces

Suppose we wish to move away from the notion of a metric space to a more general space, called a topological space. I want my new space to be one in which there is a well-defined notion of continuity. How then, should I define this new space? Well, why not make use of (3) in Theorem 1? This gives that the only notion required for continuity is that inverse images of open sets are open. Thus, it makes sense for our new space to have some sense of an open set. We then ask: How do open sets behave in $ \mathbb{R}$? Let $\mathscr{O}$ be the collection of all open subsets of $\mathbb{R}$. Then, the following can be easily checked:

$\emptyset$, $\mathbb{R} \in \mathscr{O}$.

For a finite index set $I$, and open sets $\{V_i\}_{i \in I}$, $$ \bigcap \limits_{i \in I} V_i \in \mathscr{O}.$$

For an arbitrary index set $J$, and open sets $\{U_j\}_{j \in J}$, $$\bigcup \limits_{j \in J} V_j \in \mathscr{O}$$

Ahh, you say. Why not then define a space that behaves in this manner? A collection of sets that we will call open sets that behave in the same way $\mathscr{O}$ does. Well, this is precisely the motivation behind the definition of a topology.

Let $X$ be a set. We define a topology on $X$ to be a collection of sets $\mathscr{T}$ whose members we will call open sets, satisfying:

$\emptyset$, $ X \in \mathscr{T}$.

For a finite index set $I$, and open sets $\{V_i\}_{i \in I}$, $$ \bigcap \limits_{i \in I} V_i \in \mathscr{T}.$$

For an arbitrary index set $J$, and open sets $ \{U_j\}_{j \in J}$, $$ \bigcup \limits_{j \in J} U_j \in \mathscr{T}$$ We call $$ (X,\mathscr{T}) $$ a topological space.

This gives a natural definition of continuity in these spaces:

Definition: Let $(X,\mathscr{T})$ and $(Y, \mathscr{V})$ be topological spaces. Let $ f: (X, \mathscr{T}) \to (Y, \mathscr{V})$. We say that $f$ is continuous at $ a \in X$ if, for every open neighborhood $ V \in \mathscr{V}$ of $ f(a)$, there exists an open neighborhood $ U \in \mathscr{T}$ of $a$ so that $ f(x) \in V$ whenever $ x \in U$. If $ f$ is continuous at every $ a \in X$, we say that $ f$ is continuous on $ X$.

$\endgroup$
1
$\begingroup$

Note that we will require that $(X,d_X), (Y,d_Y)$ to be metric spaces and be endowed with the metric topology. This allows us to consider open sets in $X$ and $Y$ to be unions of open balls.

Now, we say that $f:X\to Y$ is continuous at $x_0$ if for any $V\subset Y$ an open neighborhood of $f(x_0)$, $f^{-1}(V)$ is open in $X$. Let's translate this. Since $Y$ has a neighborhood basis of each point as open balls $B(f(x_0),\epsilon)$ for all $\epsilon>0$ (each $B(f(x_0),\epsilon)$ is an open neighborhood of $f(x_0)$ and for any open neighborhood $V$ of $f(x_0)$, we can find an $\epsilon>0$ depending on $V$ such that $B(f(x_0),\epsilon)\subset V$). So it suffices to consider $B(f(x_0),\epsilon)=V$. Then consider $f^{-1}(V)$. This is open by assuming that $f$ is continuous. Moreover, $x_0\in f^{-1}(V)$. So since $f^{-1}(V)$ is open, there is $\delta>0$ such that $B(x_0,\delta)\subset f^{-1}(V)$. Writing this another way, for all $\epsilon>0$, there is $\delta>0$ such that for $d_X(x_0,y)<\delta$ we have $d_Y(f(x_0),f(y))<\epsilon$.

$\endgroup$
0
$\begingroup$

in the first one(metric spaces) $B_{\varepsilon}(x_0)$s are basis for topology.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.