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Let f:D -> R be a continuous function. Let {x_{n}} be a sequence in D that converges to x in D. Then f(X_{n}) -> f(x)

G-X

New member
Jul 11, 2020
21
Let \(\displaystyle f\) : \(\displaystyle D\) -> \(\displaystyle R\) be a continuous function, where \(\displaystyle D \subset R^{k}\). Let \(\displaystyle \{x_{n}\}_{n = d}^{\infty}\) be a sequence in \(\displaystyle D\) that converges to \(\displaystyle x \in D\). Then \(\displaystyle f(X_{n})\) -> \(\displaystyle f(x)\)



We need to show that \(\displaystyle f(X_{n})\) -> \(\displaystyle f(x)\) or \(\displaystyle lim_{x_{n} -> \infty}f(x_{n}) = f(x)\).

We know that f is continuous at \(\displaystyle a \in D\). We also know that \(\displaystyle f\) is

continuous at \(\displaystyle a\), by definition, \(\displaystyle \forall z \in B_{\delta}(a) \cap D\) s.t. \(\displaystyle |f(z) - f(a)| < \epsilon\).

Since, \(\displaystyle \{x_{n}\}_{n = d}^{\infty}\) is a sequence in \(\displaystyle D\) that converges to \(\displaystyle x \in D\) then there

\(\displaystyle \exists N_{\epsilon}\) such that \(\displaystyle n \ge N_{\epsilon}\) then \(\displaystyle |f(x) - f(x_{n})| < \epsilon\). Let \(\displaystyle a = x \in D\), thus

we can conclude that \(\displaystyle |f(x) - f(x_{n})| = |f(z) - f(x)| < \epsilon\). We now need to

show that \(\displaystyle \exists z = x_{n} \in B_{\delta}(x) \cap D\). By definition of convergence, we know

\(\displaystyle \forall \epsilon > 0\) that the \(\displaystyle d(x, x_{n}) < \epsilon\). By definition of continuous, we know that

\(\displaystyle B_{\delta}(x) \cap D \neq \varnothing\). By definition of neighborhood, \(\displaystyle N(x, \delta) =

\) \{\(\displaystyle y \in R : d(x, y) < \delta\}\). This implies that \(\displaystyle \exists \epsilon = \delta\) such that at least a single

\(\displaystyle y \in B_{\delta}(x) \cap D\). Since the sequence is bounded by \(\displaystyle x_{n}\) this implies \(\displaystyle \exists z = x_{n} = y\).

We have shown that \(\displaystyle |f(x) - f(x_{n})| = |f(z) - f(x)| < \epsilon\).
 

G-X

New member
Jul 11, 2020
21
We need to show that \(\displaystyle lim_{n -> \infty}f(x_{n}) = f(x)\).

Since, \(\displaystyle \{x_{n}\}_{n = d}^{\infty}\) is a sequence in \(\displaystyle D\) that converges to \(\displaystyle x \in D\) then there

\(\displaystyle \exists N_{\epsilon}\) such that \(\displaystyle n \ge N_{\epsilon}\) then \(\displaystyle |x - x_{n}| = r < \epsilon_{1}\). Since

\(\displaystyle \delta > 0, ((B_{\delta}(x) \cap D) / \{x\}) \neq \varnothing\). We know that there \(\displaystyle \exists \epsilon_{1} = \delta\) meaning that

\(\displaystyle |x - x_{n}| = r < \delta\), so \(\displaystyle \exists z = x_{n} \in B_{\delta}(x) \cap D\).

Let \(\displaystyle a = x \in D\), thus \(\displaystyle f\) is continuous at \(\displaystyle x\), by definition,

\(\displaystyle \forall z \in B_{\delta}(x) \cap D\) s.t. \(\displaystyle |f(z) - f(x)| < \epsilon_{2}\). We know that \(\displaystyle \exists z = x_{n} \in B_{\delta}(x) \cap D\),

so we can conclude that \(\displaystyle |f(x_{n}) - f(x)| < \epsilon{2}\).

Thus we have proven that \(\displaystyle lim_{n -> \infty}f(x_{n}) = f(x)\).
 

G-X

New member
Jul 11, 2020
21
We need to show that \(\displaystyle lim_{n -> \infty}f(x_{n}) = f(x)\).

By definition of convergence, \(\displaystyle |x - x_{n}| < \epsilon_{1}\). Since \(\displaystyle \delta > 0\), this implies that \(\displaystyle ((B_{\delta}(x) \cap D) / \{x\}) \neq \varnothing\).

We know that by definition of convergence this statement holds \(\displaystyle \forall \epsilon_{1} > 0\), so there \(\displaystyle \exists \epsilon_{1} = \delta\) implying that \(\displaystyle |x - x_{n}| < \delta\).

If the \(\displaystyle d(x, x_{n}) < \delta\) then \(\displaystyle x_{n} \in B_{\delta}(x)\). We also know that \(\displaystyle x_{n} \in D\), thus we can conclude that \(\displaystyle x_{n} \in B_{\delta}(x) \cap D\).

Let \(\displaystyle a = x \in D\), thus \(\displaystyle f\) is continuous at \(\displaystyle x\), by definition, \(\displaystyle \forall z \in B_{\delta}(x) \cap D\) s.t. \(\displaystyle |f(z) - f(x)| < \epsilon_{2}\).

We know that \(\displaystyle \exists z = x_{n} \in B_{\delta}(x) \cap D\), so we can conclude that \(\displaystyle |f(x_{n}) - f(x)| < \epsilon{2}\), \(\displaystyle \forall \epsilon_{2} > 0\) by definition of continuous.

Thus we have proven that \(\displaystyle lim_{n -> \infty}f(x_{n}) = f(x)\).
 
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