Continuity Theorem and Converging Sequences

If a function $ f: X \to Y $ is continuous and a sequence of points $ x_1, x_2, \dots $ in $ X $ converges to a point $ x $, then the sequence of values $ f(x_1), f(x_2), \dots $ will converge to $ f(x) $ in $ Y $.

In other words, a continuous function preserves the convergence of sequences.

If the points $ x_n $ move closer and closer to $ x $, then the values of the function $ f(x_n) $ will similarly approach $ f(x) $.

A practical example
The proof

A practical example

Let’s take the function $ f: \mathbb{R} \to \mathbb{R} $, defined as $ f(x) = 2x $, and the sequence $ x_n = \frac{1}{n} $, where $ n \in \mathbb{N} $.

This sequence $ (x_n) $ converges to $ 0 $ as $ n \to \infty $.

The terms of the sequence are $ x_1 = 1 $, $ x_2 = \frac{1}{2} $, $ x_3 = \frac{1}{3} $, and so on.

It’s clear that $ x_n $ approaches $ 0 $ as $ n $ increases.

Now, let's apply the function $ f $ to each term of the sequence:

$$ f(x_1) = f(1) = 2 $$

$$ f(x_2) = f\left(\frac{1}{2}\right) = 1 $$

$$ f(x_3) = f\left(\frac{1}{3}\right) = \frac{2}{3} $$

$$ ... $$

The new sequence $ f(x_n) = 2x_n $ is $ 2, 1, \frac{2}{3}, \dots $, which converges to $ 0 $.

So, the sequence $ (f(x_n)) $ converges to $ f(0) = 0 $, just as the theorem suggests.

Thus, the continuous function has preserved the convergence of the sequence.

The proof

To prove that $ (f(x_n)) $ converges to $ f(x) $, we need to start from the assumption that $ f $ is continuous and use this property.

Continuity implies that the inverse image of an open set in $ Y $ is also an open set in $ X $.

We’ll use this property to show that, given any neighborhood $ U $ of $ f(x) $, the terms of the sequence $ f(x_n) $ eventually lie within $ U $ for sufficiently large $ n $.

Step 1: Arbitrary neighborhood of $ f(x) $

Consider an arbitrary neighborhood $ U $ of $ f(x) $ in $ Y $.

This means that $ U $ is an open set containing $ f(x) $.

The goal is to prove that, for sufficiently large $ n $, the terms of the sequence $ f(x_n) $ will be contained within $ U $.

Step 2: Preimage of $ U $ under $ f $

Since $ f $ is continuous, we know that the inverse image of $ U $ under $ f $, denoted by $ f^{-1}(U) $, is an open set in $ X $.

This means that every point $ x $ in $ f^{-1}(U) $ has its image under $ f $ in $ U $.

Furthermore, since $ f(x) \in U $, we know that $ x \in f^{-1}(U) $.

Step 3: Sequence $ (x_n) $ converging to $ x $

By assumption, the sequence $ (x_1, x_2, \dots ) $ in $ X $ converges to $ x $.

This means that, for every neighborhood of $ x $, there exists an index $ N $ such that, for all $ n \geq N $, the terms $ x_n $ of the sequence are contained within that neighborhood.

Since $ f^{-1}(U) $ is an open neighborhood of $ x $, we can apply this definition of convergence to the sequence $ (x_n) $.

Step 4: Existence of $ N $

Since $ (x_n) $ converges to $ x $, we know that there exists a natural number $ N $ such that for all $ n \geq N $, the terms $ x_n $ are contained within $ f^{-1}(U) $.

This implies that for all $ n \geq N $, we have $ x_n \in f^{-1}(U) $, or equivalently, $ f(x_n) \in U $.

Conclusion

Since $ f(x_n) \in U $ for all $ n \geq N $, we can conclude that the sequence $ (f(x_n)) $ converges to $ f(x) $.

This proves that a continuous function $ f $ preserves the convergence of sequences: if $ (x_n) $ converges to $ x $, then $ (f(x_n)) $ converges to $ f(x) $.

And so on.